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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