Mccord (Pmccord) – HW6 Acids, Bases and Salts – Mccord – (51520) 1 This Print-Out Should Have 45 Questions

Home , Barium hydroxide

mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 1 This print-out should have 45 questions. 6. The term is misleading, because the am- Multiple-choice questions may continue on monium ion is not an acid. the next column or page – find all choices Explanation: before answering. 003 10.0 points 001 10.0 points −9 If the value of Kb for pyridine is 1.8 × 10 , Assume that five weak acids, identified only calculate the equilibrium constant for by numbers (1, 2, 3, 4, and 5), have the + C5H5NH (aq)+ H2O(ℓ) → following ionization constants. + C5H5N(aq) + H3O (aq) .

− Ionization 1. 5.6 × 10 6 correct Acid Constant − Ka value 2. 1.8 × 10 9 − 1 1.0 × 10 3 − × −16 2 3.0 × 10 5 3. 1.8 10 − 3 2.6 × 10 7 − × 8 4 4.0 × 10 9 4. 5.6 10 −11 5 7.3 × 10 − 5. −1.8 × 10 9 The anion of which acid is the strongest Explanation: base? 004 10.0 points 1. 4 Which of the following is true in pure water at any temperature? 2. 5 correct 1. Kw decreases with increasing tempera- 3. 2 ture.

+ − −14 4. 3 2. [H3O ][OH ] = 1.0 × 10

+ − 5. 1 3. [H3O ] = [OH ] correct Explanation: 4. pH = 7.0 or greater than 7.0 002 10.0 points 5. pH = 7.0 The term “Ka for the ammonium ion” de- scribes the equilibrium constant for which of Explanation: the following reactions? Kw is shown to INCREASE with increasing temperature. pH = 7 is only true when + ⇀ + ◦ + − 1. NH4 +H2O ↽ NH3 +H3O correct water is at 24 C. [H3O ][OH ] = Kw, which increases with temperature. + − 2. NH3 +H2O ↽⇀ NH4 + OH At high temperatures pH can be less than + − 7. Thus [H3O ] = [OH ] is the only case that + ⇀ + 3. NH3 +H3O ↽ NH4 +H2O is true. + − ⇀ 4. NH4 + OH ↽ NH3 +H2O 005 10.0 points Which is NOT a conjugate acid-base pair? + − 5. NH4Cl(solid) + H2O ↽⇀ NH + Cl 4 − 1. H2O:OH mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 2

+ Kw − [H O ] = 2. HCl : Cl 3 [OH−] 1.0 × 1014 + × −6 3. H SO : H SO = − =3.0 10 M 3 4 2 4 3.3 × 10 9 − 4. H : H 2 008 10.0 points − 2− What is [OH ] in a 0.0050 M HCl solution? 5. H2SO4 : SO4 correct − Explanation: 1. 6.6 × 10 5 M 2− Except for H2SO4 and SO4 , the members − of all of the pairs diﬀer by one proton. 2. 5.0 × 10 3 M

− 006 10.0 points 3. 1.0 × 10 7 M − What is the conjugate acid of NO ? 3 − 4. 2.0 × 10 12 M correct − 1. NO2 5. 1.0 M 2. NH3 Explanation: − 3. H+ [OH ]=0.0050 M Since HCl is a strong acid, it completely dissociates and H+ is 0.0050 M. 4. HNO3 correct − 2− HCl ↽⇀ H+ + Cl 5. NO3

− 6. OH + − × −14 Explanation: Kw = [H ][OH ]=1 10 − K Since the question asks for the conjugate [OH ] = w − [H+] acid, we can assume NO3 is acting as a base. −14 This means that it is a proton acceptor. To 1 × 10 − = =2 × 10 12 M form the conjugate acid, it accepts a H making 0.0050 HNO3.

007 10.0 points 009 10.0 points + − × −9 Which pH represents a solution with 1000 What is [H3O ] when [OH ]=3.3 10 M? − times higher [OH ] than a solution with pH − 1. 1.0 × 10 7 M of 5?

− 2. 3.3 × 10 9 M 1. pH=2

− 3. 3.3 × 10 5 M 2. pH = 0.005

− 4. 3.0 × 10 6 M correct 3. pH=8 correct

− 5. 6.6 × 10 5 M 4. pH=1 Explanation: − − 5. pH=3 [OH ]=3.3 × 10 9 M + − 14 Kw = [H3O ][OH ]=1 × 10 6. pH = 5000 mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 3 Hydroxylamine is a weak molecular base with −9 7. pH=7 Kb =6.6 × 10 . What is the pH of a 0.0500 M solution of hydroxylamine? 8. pH=4 1. pH = 8.93 9. pH=6 Explanation: 2. pH = 7.12 pH=5 3. pH = 3.63 pOH = 14 − pH = 14 − 5=9 4. pH = 4.74 − − − [OH ]=10 pOH = 10 9 M 5. pH = 9.26 correct − − 3 −9 [OH ]x = 1000 [OH ] = (10 )(10 M) − 6. pH = 9.48 = 10 6 M 7. pH = 10.37 pOHx = − log(OHx)=6

pHx = 14 − pOHx = 14 − 6=8 Explanation: Hydroxylamine is a weak base, so use the − equation to calculate weak base [OH ] con- 010 10.0 points centration (note that this is the approximate What isthe pH of a 0.12 M Ba(OH)2 aqueous equation. Why? Because Kb is very small solution? and the concentration is reasonable) : − [OH ] = K C 1. 1.33802 b b = p(6.6 × 10−9)(0.0500) 2. 8.7 − =1q.82 × 10 5 − 3. 0.619789 After finding [OH ], you can find pH using either method below: A) 4. 13.3802 correct − pOH = − log 1.82 × 10 5 =4.74 5. 10.0352 − pH = 14 4.74=9.26 Explanation: or B) [Ba(OH) ]=0.15 M + Kw 2 [H ] = − Ba(OH)2 is a strong base which dissociates [OH ] × −14 in aqueous solution to produce two moles of 1.0 10 × −10 − = − =5.52 10 OH for every mole of Ba(OH)2, so 0.12 M 1.82 × 10 5 − − Ba(OH)2 produces 0.24 M OH . pH = − log 5.52 × 10 10 =9.26 2+ − Ba(OH)2 → Ba + 2OH ini 0.12M 0M 0M 012 10.0 points ∆ −0.12M +0.12M 2(0.12 M) What is the pH of a 0.2 M solution of potassium generate (KR-COO)? Ka for the generic fin 0 M +0.12M +0.24 M − acid (R-COOH) is 2.7 × 10 8. pH = 14−pOH = 14−(− log0.24) = 13.3802 1. 10.285

2. 7.000 011 10.0 points mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 4 3. 10.565 Your answer must be within ± 0.4% Correct answer: 2.70956. 4. 10.195 Explanation: − M =0.16 M K =4.2 × 10 10 5. 3.565 C6H5NH3NO3 b It’s a salt of a weak base (BHX).This means you need a K for the weak acid BH+: 6. 7.569 a Kw 7. 6.431 Ka = Kb 1.0 × 10−14 8. 3.435 = 4.2 × 10−10 −5 9. 10.435 correct =2.38095 × 10

10. 10.805 You CAN use the approximation for the equilibrium which means that Explanation: −8 MKR−COO =0.2 M Ka =2.7 × 10 + [H ] = Ka · CBH+ It’s a salt of a weak generic acid (KA). Get it? Generic acid makes generic ions. = p(2.38095 × 10−5)(0.16) Ha! This means you need a Kb for the weak =0q.0019518 M − Kw base A . Use Kb = and you’ll get the Ka pH = − log(0.0019518) = 2.70956 −7 Kb = 3.7037 × 10 . You CAN use the approximation for the equilibrium which means that 015 10.0 points − The pH of lemon juice is approximately 2.4. · − [OH ] = Kb CA =0.000272166 M At this pH, the hydronium ion concentration p is closest to which value? pH = 14 − pOH −12 = 14+ log(0.000272166) = 10.4348 1. 2.50 × 10 M − 2. 5.62 × 10 4 M 013 10.0 points ◦ − At 25 C, the pH of a water solution of a salt 3. 4.00 × 10 3 M correct of a WEAK acid and a STRONG base is 4. 250 M 1. less than 7. Explanation: pH = 2.4, so 2. greater than 7. correct −2.4 M + = 10 =0.00398107 M 3. about 7. H

4. equal to the hydrogen ion concentration. 016 10.0 points Explanation: Which solution has the highest pH?

014 10.0 points 1. 0.1 M of KHCOO, −4 What is the pH of a 0.16 M solution of anilin- Ka HCOOH =1.8 × 10 ium nitrate (C6H5NH3NO3)? Kb for aniline −10 is 4.2 × 10 . 2. 0.1 M of KCl, Ka HCl = very large mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 5

3. 0.1 M of KCH3COO, 019 10.0 points × −5 Ka HC2H3O2 =1.8 10 A solution has a pH of 4.35. Find the pOH.

× −4 4. 0.1 M of KNO2, Ka HNO2 =4.5 10 1. 4.35

−8 5. 0.1 M of KClO, Ka HClO = 3.5 × 10 2. 9.65 correct correct 3. None of these Explanation: 4. 18.35 017 10.0 points What is the pH of a solution that contains Explanation: 11.7 g of NaCl for every 200 mL of solution? pH = 4.35

1. 1.0 pOH = 14 − pH=9.65

− 2. 10 1 020 (part 1 of 2) 10.0 points 3. 7.0 correct The pH of an aqueous solution is measured as + 1.21. Calculate the [H3O ]. − 4. 1.0 × 10 7 Correct answer: 0.0616595 M. Explanation: m =11.7 g V = 200 mL Explanation: NaCl soln + NaCl completely dissociates in water to give pH=1.21 [H3O ] = ? − Na+ and Cl , neither of which hydrolyzes and + − pH = − log[H O+] so in aqueous NaCl the H3O and OH ions 3 + result from autoionization of water. log[H3O ] = −pH + [H3O ] = antilog(−pH) + − −14 × −pH Kw = [H3O ][OH ]=1 × 10 =1 10 + − −7 × −1.21 [H3O ] = [OH ]=1 × 10 =1 10 =0.0616595 M + and pH = − log[H3O ]=7.0 021 (part 2 of 2) 10.0 points − 018 10.0 points Calculate the [OH ]. A 0.010 M solution of a weak acid HA has a − pH of 4.20. What is the pOH of the solution? Correct answer: 1.62181 × 10 13 M. Explanation: 1. 14.0 − pH=1.21 [OH ] = ? 2. None of these + − × −14 2 3. 4.20 [H3O ][OH ]=1 10 M −14 2 − 1 × 10 M [OH ] = + 4. 7.0 [H3O ] 1 × 10−14 M2 = 5. 9.80 correct 0.0616595 M − Explanation: =1.62181 × 10 13 M mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 6 dissolving 0.100 mole of sodium acetate 022 10.0 points (NaCH3COO) in enough water to make one What is the pH of a solution made by mixing liter of solution? Ka for CH3COOH is − 0.05 mol of NaCN with enough water to make 1.80 × 10 5. a liter of solution? −10 Ka for HCN is 4.9 × 10 and Kw = 1. 8.87 correct − 1 × 10 14. 2. 9.25 Correct answer: 11.0044. 3. 5.13 Explanation: − 4. 5.56 × 10 11 − [OH ] = Kb C 5. 10.25 K = p w C K r a 6. 5.74 × −14 1 10 −10 = − (0.05) = 0.00101015 7. 5.56 × 10 s4.9 × 10 10 − 8. 1.80 × 10 6 K [H+] = w × −6 [OH−] 9. 7.46 10 −14 1 × 10 − −9 = =9.89949 × 10 12 10. 1.34 × 10 0.00101015 Explanation:

pH = − log[H+] 025 10.0 points − = − log(9.89949 × 10 12)=11.0044 A 0.200 M solution of a weak monoprotic acid HA is found to have a pH of 3.00 at room temperature. What is the ionization constant 023 10.0 points of this acid? Identify the list in which all salts produce a × −3 basic aqueous solution. 1. 5.0 10 − 2. 2.0×10 5 1. AgNO3, NaCHO2, CrI3 − 3. 1.0×10 6 2. NH4Cl, C6H4NH3NO3, FeI3 4. 5.30 3. AlCl3, Zn(NO3)2, KClO4 − 5. 5.0×10 6 correct 4. CH3NH3Cl, KNO3, NaBz (sodium ben- zoate) − 6. 1.8×10 5

5. KCH3COO, NaCN, KF correct − 7. 2.0×10 9 Explanation: − 8. 1.0×10 3 024 10.0 points What is the pH in a solution made by Explanation: mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 7

026 10.0 points 4. 0.21% What is the percent ionization for a weak acid −7 HX that is 0.40 M? Ka =4.0 × 10 . 5. 0.69% Explanation: 1. 0.00020% 029 10.0 points 2. 0.050% A 20 mL sample of 0.20 M nitric acid solution is required to neutralize 40 mL of barium 3. 0.020% hydroxide solution. What is the molarity of the barium hydroxide solution? 4. 0.10% correct 1. 0.050 M correct 5. 2.0% Explanation: 2. 0.025 M

027 10.0 points 3. 0.100 M A 0.28 M solution of a weak acid is 3.5% ionized. What is the pH of the solution? 4. 0.0025 M

1. 2.01 correct 5. 0.200 M Explanation: 2. 1.46 VHNO3 =20mL [HNO3] = 0.20M

VBa(OH)2 = 40 mL 3. 5.25 The balanced equation for this neutralization reaction is 4. 0.55 2HNO + Ba(OH) → Ba(NO ) +2H O 5. 3.17 3 2 3 2 2 We determine the moles of HNO used: Explanation: 3 M =0.28 M P =3.5% ? mol HNO3 =0.020 L soln 3.5% of the 0.28 M is ionized (contributes × 0.20 mol HNO3 to pH), so 1 L soln =0.0040 molHNO3 3.5 Using the mole ratio from the chemical equa- [H+]=(0.28 M) × =0.0098 M 100 tion we calculate the moles Ba(OH)2 needed to react with 0.0040 mol of HNO3: pH = − log[H+] = − log(0.0098) = 2.00877 ? mol Ba(OH)2 =0.0040 mol HNO3 × 1 mol Ba(OH)2 028 10.0 points 2 mol HNO3 The pH of 0.010 M aniline(aq) is 8.32. =0.0020 mol Ba(OH) What is the percentage aniline protonated? 2 There are 0.0020 moles Ba(OH)2 in the 40 mL 1. 2.1% sample. Molarity is moles solute per liter of solution: 0.0020 moles Ba(OH) 2. 0.021% correct ? M Ba(OH) = 2 2 0.040 L solution 3. 0.12% =0.050 MBa(OH)2 mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 8

030 10.0 points 032 10.0 points When an acid and base neutralize each other, A 29.1 mL sample of a solution of RbOH is the products are generally water neutralized by 22.51mLofa2.735 M solution of HBr. What is the molarity of the RbOH 1. a salt. correct solution?

2. a gel. Correct answer: 2.11563 M.

3. a colloid. Explanation: VRbOH = 29.1 mL VHBr = 22.51 mL 4. an ion. [HBr] = 2.735 M [RbOH] = ? Explanation: The general format for neutralization reac- −→ tions is acid + base → salt + water. RbOH + HBr RbBr + H2O

2.735mol HBr L 031 10.0 points (22.51 mL) L 1000 mL How many moles of Ca(OH)2 are needed to neutralize three moles of HCl? =0.0615649 mol HBr

1. three 1 mol RbOH (0.0615649 mol HBr) 1 mol HBr 2. 1.5 correct × 1 1000 mL 3. four 29.1 mL L mol =2.11563 RbOH 4. eight L =2.11563 M RbOH 5. 0.5 033 10.0 points For the neutralization reaction involving 6. two HNO3 and LiOH, how much of 2.10 M HNO3 is needed to neutralize 22.2 L of a 4.66 M 7. six LiOH solution? The molar mass of LiOH is 23.95 g/mol. The molar mass of HNO3 is 63.1 8. one g/mol. The density of the HNO3 solution is Explanation: 1.06 g/mL. The density of the LiOH solution nHCl = 3 mol is 1.15 g/mL. For acid base neutralization we need one − mole of H+ for every mole of OH . Therefore 1. 0.567 g the balanced equation is 2. 109.7 g → 2HCl + Ca(OH)2 CaCl2 +2H2O 3. 56,600 g

4. 56.6 g ? mol Ca(OH)2 = 3 mol HCl × 1 mol Ca(OH)2 5. 52,200 g correct 2 mol HCl =1.5 mol Ca(OH)2 6. 103.5 g mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 9 acid that is 1.3 percent ionized or dissociated. 7. 49.3 g What is the pH?

8. 1,620,000 g 1. 2.3 correct

Explanation: 2. 0.3 The reaction is → HNO3(aq) + LiOH(aq) 3. 0.4 LiNO3(aq)+ H2O(ℓ) Find the number of moles of LiOH used: 4.66 mol LiOH 4. 1.5 (22.2 L LiOH) × 1 L LiOH = 103.452 mol LiOH 5. 4.3 Find the moles of HNO3 needed: Explanation: 1 mol HNO3 (103.452 mol LiOH) × [CH3COOH] = 0.4M percent=1.3% 1 mol LiOH First calculate the concentration of acetic = 103.452 mol HNO 3 acid that is ionized: Finally, ﬁnd the mass of HNO3: 1 L HNO3 − + (103.452 mol HNO3) × CH3COOH → CH3COO +H 2.1 mol HNO3 ×1.06 g HNO3 × 1000 mL 1.3 0.4 M × =0.0052 M H+ . 1 mL HNO3 1 L 100 = 52218.6 g HNO 3 Thus

034 10.0 points pH = − log H+ = − log (0.0052) = 2.284 . An aqueous solution is prepared with 2 moles of HCl and 1 mole of Ca(OH)2. The resulting solution contains mainly of 036 10.0 points − Determine the total ionic equation for the 1. water and Cl ,H+, and Ca2+ ions. reaction between HBr(aq) and Ba(OH)2(aq). − 2. water and Cl and Ca2+ ions. correct + − 1. 2H +2OH → 2H2O − + − 2+ 3. water and Cl , H , OH , and Ca − 2+ 2. 2Br + Ba → BaBr2 ions.

− − 3. 2HBr + Ba(OH)2 → BaBr2 +2H2O 4. water and Cl , OH , and Ca2+ ions. − − 4. 2H+ +2Br + Ba2+ +2OH → Explanation: − Ba2+ +2Br +2H O correct 2HCl(aq) + Ca(OH)2(aq) → 2 CaCl2(aq)+2H2O(aq) Explanation: 1 mole of Ca(OH)2 reacts with 2 moles of An acid and base react to produce a salt HCl, so there will be no Ca(OH)2 nor HCl and water. In this case the salt, formed from 2+ left. The CaCl2(aq) will exist as Ca (aq) − the available cations and anions other than and Cl (aq). The H+ from the HCl and + − − H and OH , is barium bromide (BaBr2), the OH from the Ca(OH)2 have all reacted. which is soluble. The formula unit equation is + − Only a miniscule amount of H and OH → remain from the autoionization of the water. Ba(OH)2(aq) + 2HBr(aq) BaBr2(aq)+2H2O(ℓ). 035 10.0 points In the total ionic equation soluble compounds Assume you have a 0.4 M solution of acetic are written as their ions: mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 10

2+ − + − → Ba(aq) +2OH(aq) +2 H(aq) + Br(aq) 2. Sr(NO3)2 +2H2O correct h i2+ h 1− i Ba(aq) +2Br(aq) +2H2O(ℓ) h i 3. SrNO3 +H2O 037 10.0 points If aqueous acetic acid is reacted with sodium 4. SrH2 + HNO5 hydroxide, which of the following substances Explanation: are in the net ionic equation? Sr(OH)2 is a base and HNO3 an acid; they create a salt and water. 1. acetate ion, hydroxide ion, hydronium ion, and water 040 10.0 points 2. acetate ion, hydronium ion, and water Aqueous ammonia can be used to neutralize sulfuric acid (H2SO4) and nitric acid (HNO3) 3. acetic acid, hydroxide ion, acetate ion, to produce two salts extensively used as fer- and water correct tilizers. They are

4. acetic acid, hydroxide ion, hydronium ion, 1. (NH4)2SO4 and NH4NO3, respectively. acetate ion, and water correct

5. acetic acid, sodium ion, hydroxide ion, 2. NH4SO4 and NH4NO3, respectively. and acetate ion 3. NH4SO3 and NH4OH, respectively. Explanation: 4. cyanamide and cellulose nitrate, respec- 038 10.0 points tively. Identify the products of the chemical equation Explanation: Aqueous ammonia is a weak base which 3LiOH+H PO → 3 4 reacts with acids to form salts. With sulfuric acid and nitric acid, it forms ammonium 1. 3LiH + (OH)3PO4 sulfate and ammonium nitrate, respectively, both of which are used as fertilizers. 2. Li3PO4 +3H2O correct 041 10.0 points 3. 3H+3O2 +H3Li3 Identify the salt that is produced from 4. Li3P+2H2O+H3O5 the acid-base neutralization reaction be- Explanation: tween potassium hydroxide and acetic acid (CH3COOH). Acid + Base → Salt + Water 1. potassium cyanide

039 10.0 points 2. potassium acetate correct What are the products of the following reaction? 3. potassium formate

Sr(OH)2 + 2HNO3 → 4. potassium amide Explanation: 1. Sr(NO2)2 +2H2O2 The balanced equation is mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 11

+ − CH3CO2H(aq) + K (aq) + OH (aq) → It was found that 25 mL of 0.012 M HCl + − K (aq) + CH3CO2 (aq)+ H2O(ℓ) neutralized 40 mL of NaOH solution. What Potassium acetate (CH3CO2K) is the salt. was the molarity of the base solution?

042 10.0 points 1. 0.006 M What volume of 0.585 M Ca(OH)2 would be needed to neutralize 15.8 L of 1.51 M HCl? 2. 0.012 M

1. 40.8 L 3. 0.050 M

2. 12.2 L 4. 0.0075 M correct Explanation: 3. 6.12 L VHCl = 25 mL MHCl =0.012 M VNaOH =40mL=0.04 L 4. 3.06 L The base is NaOH. To neutralize, mol H+ − = mol OH . 5. 20.4 L correct Explanation: 0.012 mol [Ca(OH)2] = 0.585 M VHCl = 15.8 L nH+ = (25 mL HCl) [HCl] = 1.51 M L + The balanced equation for this neutraliza- × 1 L × 1 mol H tion reaction is 1000 mL 1 mol HCl + =0.0003 mol H = nOH− = nNaOH 2HCl + Ca(OH)2 → CaCl2 +2H2O We determine the moles of HCl present: mol 0.0003 mol NaOH 1.51 mol HCl M = = ? mol HCl = 15.8 L soln × NaOH L 0.04 L 1 L soln =0.0075 M NaOH = 23.86 mol HCl Using the mole ratio from the chemical 044 10.0 points equation we calculate the moles of Ca(OH)2 needed to react with this amount of HCl: The pH of a solution of hydrochloric acid is 1.57. What is the molarity of the acid? ? mol Ca(OH)2 = 23.86 mol HCl Correct answer: 0.0269 mol/L. × 1 mol Ca(OH)2 2 mol HCl Explanation: = 11.93 mol Ca(OH)2 045 10.0 points We use the molarity of the Ca(OH) so- 2 How many moles of NaOH are needed to neu- lution to convert from moles to volume of tralize three moles of HCl? Ca(OH)2:

? L Ca(OH)2 = 11.93 mol Ca(OH)2 1. 0.5 1 L soln × 2. one 0.585 mol Ca(OH)2 = 20.4 L Ca(OH)2 3. six

043 10.0 points 4. 1.5 mccord (pmccord) – HW6 Acids, Bases and Salts – mccord – (51520) 12

5. three correct

6. two

7. eight

8. four Explanation: For acid base neutralization we need one − mole of H+ for every mole of OH . Therefore the balanced equation is

HCl + NaOH → NaCl+ H2O 1 mol NaOH ? mol NaOH = 3 mol HCl × 1 mol HCl = 3 mol NaOH