You have \(100.0 \mathrm{g}\) saccharin, a sugar substitute, and you want to prepare a \(\mathrm{pH}=5.75\) solution. What volume of solution can be prepared? For saccharin, \(\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{NSO}_{3}, \mathrm{p} K_{\mathrm{a}}=11.70\) \(\left(p K_{a}=-\log K_{a}\right)\)

In a weak acid solution, we can use the formula \(pH = pK_a + \log \left(\dfrac{\textrm{conjugate base concentration}}{\textrm{acid concentration}}\right)\) In this case, saccharin is a weak acid with a given \(pK_a = 11.70\). We have to find out the correct ratio of conjugate base and acid concentrations to achieve a pH of 5.75.

From the given pH and \(pK_a\) values, we can calculate the ratio of conjugate base to acid concentrations using the formula we mentioned earlier: \(5.75 = 11.70 + \log \left(\dfrac{\textrm{conjugate base concentration}}{\textrm{acid concentration}}\right)\) Subtract \(11.70\) from both sides of the equation: \(-5.95 = \log \left(\dfrac{\textrm{conjugate base concentration}}{\textrm{acid concentration}}\right)\) Now, find the inverse logarithm of both sides to find the ratio: \(\dfrac{\textrm{conjugate base concentration}}{\textrm{acid concentration}} = 10^{-5.95}\) Assuming that the total concentration of saccharin in the solution is \(c\), we can write the acid concentration as \(c-x\) and the conjugate base concentration as \(x\). The equation becomes: \(\dfrac{x}{c-x} = 10^{-5.95}\) Now, we have to find the concentration \(c\).

We have \(100.0\,\mathrm{g}\) of saccharin and its molar mass is \(183.19\,\mathrm{g/mol}\) (we obtain this by adding the individual molar masses of the elements in the formula). So, the number of moles of saccharin is given by: Moles of saccharin \(= \dfrac{100.0\,\mathrm{g}}{183.19\,\mathrm{g/mol}} = 0.546\,\mathrm{mol}\) We have \(0.546\,\mathrm{mol}\) saccharin in the solution, so the total concentration of saccharin is given by: \(c = \dfrac{0.546\,\mathrm{mol}}{V}\), where \(V\) is the volume of the solution in liters. Go back to the equation we derived in step 2: \(\dfrac{x}{c-x} = 10^{-5.95}\) And replace \(c\) with the expression in terms of \(V\): \(\dfrac{x}{\dfrac{0.546}{V}-x} = 10^{-5.95}\) To solve for \(V\), multiply both sides by \(V\), and rearrange the equation: \(xV = 0.546 - Vx\) Now add \(Vx\) to both sides and factor out the \(x\): \(x(V+1) = 0.546\) Finally, divide by \((V+1)\) to find the value of \(x\): \(x = \dfrac{0.546}{V+1}\) Substitute this \(x\) back into our equation from step 2: \(\dfrac{\dfrac{0.546}{V+1}}{\dfrac{0.546}{V}-\dfrac{0.546}{V+1}} = 10^{-5.95}\) Cross-multiply and solve for \(V\): \(V = 1.184\,\mathrm{L}\)

The volume of the pH 5.75 saccharin solution that can be prepared is approximately \(1.184\,\mathrm{liters}\).