Chemistry: Atoms First (2nd Edition)

Published by Cengage Learning
ISBN 10: 1305079248
ISBN 13: 978-1-30507-924-3

Chapter 13 - Exercises - Page 575d: 82

Answer

It is necessary to have 344ml of solution to prepare one with pH = 5.75.

Work Step by Step

1. Drawing the equilibrium (ICE) table we get these concentrations at equilibrium:** The image is in the end of this answer. -$[H_3O^+] = [C_7H_4NS{O_3}^-] = x$ -$[HC_7H_4NSO_3] = [HC_7H_4NSO_3]_{initial} - x$ 2. Calculate the Ka value: $K_a = 10^{-pKa}$ $K_a = 10^{- 11.7}$ $K_a = 1.995 \times 10^{- 12}$ 3. Find the $[H_3O^+]$ value: $[H_3O^+] = 10^{-pH}$ $[H_3O^+] = 10^{- 5.75}$ $[H_3O^+] = 1.778 \times 10^{- 6}M$ Therefore : $x = 1.778 \times 10^{- 6}M$ 4. Now, use the Ka and x values and equation to find the initial concentration value. $Ka = \frac{[H_3O^+][C_7H_4NS{O_3}^-]}{ [Initial HC_7H_4NSO_3] - x}$ $ 1.995\times 10^{- 12}= \frac{[x^2]}{ [Initial HC_7H_4NSO_3] - x}$ $ 1.995\times 10^{- 12}= \frac{( 1.778\times 10^{- 6})^2}{[Initial HC_7H_4NSO_3] - 1.778\times 10^{- 6}}$ $[Initial HC_7H_4NSO_3] - 1.778\times 10^{- 6} = \frac{ 3.162\times 10^{- 12}}{ 1.995\times 10^{- 12}}$ $[Initial HC_7H_4NSO_3] - 1.778\times 10^{- 6} = 1.585$ $[Initial HC_7H_4NSO_3] = 1.585 + 1.778\times 10^{- 6}$ $[Initial HC_7H_4NSO_3] = 1.585M$ 5. Now, calculate the number of moles in 100g of saccharin: Molar Mass ($HC_7H_4NSO_3$): 1.01* 1 + 12.01* 7 + 1.01* 4 + 14.01* 1 + 32.07* 1 + 16* 3 = 183.2g/mol $n(moles) = \frac{mass(g)}{mm(g/mol)} = \frac{100}{183.2} = 0.5458$ 6. Find the volume necessary: $Concentration(M) = \frac{n(moles)}{V(L)}$ $1.585 = \frac{0.5458}{V(L)}$ $V(L) = \frac{0.5458}{1.585} = 0.344L = 344ml$ -------
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