Consider a buffer prepared from sodium bisulfate with a $K_a = 1.0 \times 10^{-2}$. Part 1 of 2: The buffer is prepared by mixing 49.6 mL of 0.052 M sodium bisulfate and 10.7 mL o... Consider a buffer prepared from sodium bisulfate with a $K_a = 1.0 \times 10^{-2}$. Part 1 of 2: The buffer is prepared by mixing 49.6 mL of 0.052 M sodium bisulfate and 10.7 mL of 0.12 M NaOH. What is the pH? Round your answer to 2 decimal places. Part 2 of 2: How many grams of HCl must be added to 57.0 mL of this buffer to change its pH by 0.050 units? Round your answer to 1 significant figure. Read more

Steps to Solve

1. Calculate moles of $HSO_4^-$ and $OH^-$

First, calculate the number of moles of sodium bisulfate ($NaHSO_4$) and sodium hydroxide ($NaOH$) used. $$ moles \ of \ NaHSO_4 = volume \times concentration = 0.0496 \ L \times 0.052 \ M = 0.0025792 \ mol $$ $$ moles \ of \ NaOH = volume \times concentration = 0.0107 \ L \times 0.12 \ M = 0.001284 \ mol $$

2. Determine the reaction and remaining moles

The $NaOH$ will react with $HSO_4^-$ to form $SO_4^{2-}$ and $H_2O$. The reaction is: $HSO_4^- + OH^- \rightarrow SO_4^{2-} + H_2O$ Since the moles of $NaOH$ are less than the moles of $NaHSO_4$, $NaOH$ will be completely consumed. The remaining moles of $HSO_4^-$ and the formed moles of $SO_4^{2-}$ are computed as follows: $$ moles \ of \ HSO_4^- \ remaining = 0.0025792 \ mol - 0.001284 \ mol = 0.0012952 \ mol $$ $$ moles \ of \ SO_4^{2-} \ formed = 0.001284 \ mol $$

3. Calculate the concentrations of $HSO_4^-$ and $SO_4^{2-}$

The total volume of the solution is $49.6 \ mL + 10.7 \ mL = 60.3 \ mL = 0.0603 \ L$. $$ [HSO_4^-] = \frac{0.0012952 \ mol}{0.0603 \ L} = 0.02148 \ M $$ $$ [SO_4^{2-}] = \frac{0.001284 \ mol}{0.0603 \ L} = 0.0213 \ M $$

4. Use the Henderson-Hasselbalch equation to find the pH

The Henderson-Hasselbalch equation is: $pH = pK_a + \log \frac{[A^-]}{[HA]}$ where $HA$ is $HSO_4^-$ and $A^-$ is $SO_4^{2-}$. First calculate the $pK_a$: $$ pK_a = -\log(K_a) = -\log(1.0 \times 10^{-2}) = 2.00 $$ Now, calculate the pH: $$ pH = 2.00 + \log \frac{0.0213}{0.02148} = 2.00 + \log(0.9916) = 2.00 - 0.0036 = 1.9964 $$ Rounding to two decimal places, the pH is 2.00.

5. Part 2: Determine the required pH change

The pH must change by 0.050 units. Since we are adding HCl, the pH will decrease. Thus, the new pH will be: $pH_{new} = 2.00 - 0.050 = 1.95$

6. Use the Henderson-Hasselbalch equation to find the new ratio of $[SO_4^{2-}]/[HSO_4^-]$

Using the Henderson-Hasselbalch equation: $1.95 = 2.00 + \log \frac{[SO_4^{2-}]}{[HSO_4^-]}$ $\log \frac{[SO_4^{2-}]}{[HSO_4^-]} = 1.95 - 2.00 = -0.05$ $\frac{[SO_4^{2-}]}{[HSO_4^-]} = 10^{-0.05} = 0.8913$

7. Set up equations for the change in moles due to $HCl$ addition

Let $x$ be the moles of $HCl$ added. The $HCl$ will react with $SO_4^{2-}$ to form $HSO_4^-$. The new moles of $SO_4^{2-}$ will be $0.001284 - x$, and the new moles of $HSO_4^-$ will be $0.0012952 + x$. The new concentrations are: $[SO_4^{2-}] = \frac{0.001284 - x}{0.057 \ L}$ $[HSO_4^-] = \frac{0.0012952 + x}{0.057 \ L}$ Using the ratio from Step 6: $\frac{[SO_4^{2-}]}{[HSO_4^-]} = \frac{0.001284 - x}{0.0012952 + x} = 0.8913$

8. Solve for x (moles of HCl)

$0.001284 - x = 0.8913 \times (0.0012952 + x)$ $0.001284 - x = 0.001154 + 0.8913x$ $0.001284 - 0.001154 = x + 0.8913x$ $0.00013 = 1.8913x$ $x = \frac{0.00013}{1.8913} = 6.873 \times 10^{-5} \ mol$

9. Calculate the mass of HCl

The molar mass of $HCl$ is $1.008 + 35.45 = 36.458 \ g/mol$. $$ mass \ of \ HCl = moles \times molar \ mass $$ $$ mass \ of \ HCl = 6.873 \times 10^{-5} \ mol \times 36.458 \ g/mol = 0.002509 \ g $$ Rounding to one significant figure, the mass of $HCl$ is $0.003 \ g$ or $3 \times 10^{-3} \ g$.