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Chapter 17: Problem 43

A 20.0-mL sample of \(0.200 \mathrm{M}\) HBr solution is titrated with \(0.200 \mathrm{M} \mathrm{NaOH}\) solution. Calculate the \(\mathrm{pH}\) of the solution after the following volumes of base have been added: (a) \(15.0 \mathrm{~mL}\) (b) \(19.9 \mathrm{~mL}\) (c) \(20.0 \mathrm{~mL},\) (d) \(20.1 \mathrm{~mL},\) (e) \(35.0 \mathrm{~mL}\)

Short Answer

Expert verified
The pH values of the solution after the given volumes of NaOH have been added are as follows: (a) 15.0 mL: pH = 1.60 (b) 19.9 mL: pH = 3.30 (c) 20.0 mL: pH = 7.00 (d) 20.1 mL: pH = 10.70 (e) 35.0 mL: pH = 12.70

Step by step solution

01

Determine the initial moles of HBr

To find the initial moles of HBr, we use the formula: Moles = Molarity × Volume where the molarity of HBr is 0.200 M and the volume is 20.0 mL (0.020 L). Moles of HBr = 0.200 M × 0.020 L = 0.004 mol
02

Calculate the moles of NaOH added for each case

We will use the same formula as in step 1 to calculate the moles of NaOH added in each case. Using molarity of NaOH = 0.200 M and given volumes: (a) Moles of NaOH added = 0.200 M × 0.015 L = 0.003 mol (b) Moles of NaOH added = 0.200 M × 0.0199 L = 0.00398 mol (c) Moles of NaOH added = 0.200 M × 0.020 L = 0.004 mol (d) Moles of NaOH added = 0.200 M × 0.0201 L = 0.00402 mol (e) Moles of NaOH added = 0.200 M × 0.035 L = 0.007 mol
03

Determine the moles of HBr and NaOH left after the reaction

For strong acid-strong base titrations, each mole of acid reacts with one mole of base, the moles left to calculate new equilibrium are as follows: (a) 0.004 mol HBr - 0.003 mol NaOH = 0.001 mol HBr left, no NaOH left (b) 0.004 mol HBr - 0.00398 mol NaOH = 0.00002 mol HBr left, no NaOH left (c) 0.004 mol HBr - 0.004 mol NaOH = no HBr left, no NaOH left (equivalence point) (d) no HBr left, 0.00402 mol NaOH - 0.004 mol HBr = 0.00002 mol NaOH left (e) no HBr left, 0.007 mol NaOH - 0.004 mol HBr = 0.003 mol NaOH left
04

Calculate the pH of the solution for each case

(a) Concentration of HBr left = 0.001 mol / (0.020 L + 0.015 L) = 0.025 M; pH = -log(0.025) = 1.60 (b) Concentration of HBr left = 0.00002 mol / (0.020 L + 0.0199 L) = 0.0005 M; pH = -log(0.0005) = 3.30 (c) At the equivalence point, the pH is neutral, so pH = 7.00 (d) Concentration of NaOH left = 0.00002 mol / (0.020 L + 0.0201 L) = 0.0005 M; pOH = -log(0.0005) = 3.30; pH = 14 - 3.30 = 10.70 (e) Concentration of NaOH left = 0.003 mol / (0.020 L + 0.035 L) = 0.05 M; pOH = -log(0.05) = 1.30; pH = 14 - 1.30 = 12.70 Now we have the pH values for all cases: (a) pH = 1.60 (b) pH = 3.30 (c) pH = 7.00 (d) pH = 10.70 (e) pH = 12.70

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Most popular questions from this chapter

A \(1.00-\mathrm{L}\) solution saturated at \(25^{\circ} \mathrm{C}\) with calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)\) contains \(0.0061 \mathrm{~g}\) of \(\mathrm{CaC}_{2} \mathrm{O}_{4} .\) Calculate the solubilityproduct constant for this salt at \(25^{\circ} \mathrm{C}\).

How many milliliters of \(0.105 \mathrm{M} \mathrm{HCl}\) are needed to titrate each of the following solutions to the equivalence point: (a) \(45.0 \mathrm{~mL}\) of \(0.0950 \mathrm{M} \mathrm{NaOH},\) (b) \(22.5 \mathrm{~mL}\) of \(0.118 \mathrm{M} \mathrm{NH}_{3},(\mathrm{c})\) \(125.0 \mathrm{~mL}\) of a solution that contains \(1.35 \mathrm{~g}\) of \(\mathrm{NaOH}\) per liter?

Explain why a mixture of \(\mathrm{CH}_{3} \mathrm{COOH}\) and \(\mathrm{CH}_{3} \mathrm{COONa}\) can act as a buffer while a mixture of \(\mathrm{HCl}\) and \(\mathrm{NaCl}\) cannot.

Calculate the pH at the equivalence point for titrating \(0.200 \mathrm{M}\) solutions of each of the following bases with \(0.200 \mathrm{M} \mathrm{HBr}\) : (a) sodium hydroxide \((\mathrm{NaOH}),(\mathbf{b})\) hydroxylamine \(\left(\mathrm{NH}_{2} \mathrm{OH}\right),(\mathbf{c})\) aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right)\).

A concentration of \(10-100\) parts per billion (by mass) of \(\mathrm{Ag}^{+}\) is an effective disinfectant in swimming pools. However, if the concentration exceeds this range, the \(\mathrm{Ag}^{+}\) can cause adverse health effects. One way to maintain an appropriate concentration of \(\mathrm{Ag}^{+}\) is to add a slightly soluble salt to the pool. Using \(K_{s p}\) values from Appendix \(D\), calculate the equilibrium concentration of \(\mathrm{Ag}^{+}\) in parts per billion that would exist in equilibrium with (a) \(\mathrm{AgCl},(\mathbf{b})\) \(\mathrm{AgBr},(\mathrm{c}) \mathrm{AgI}\)
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