Q19.31P - Chemistry: Molecular Nature Of Matter And Change

Question

A buffer is prepared by mixing $\mathbf{\text{204 mL}}$ of $\mathbf{\text{0.452 M HCl}}$ and $\mathbf{\text{0.500 L}}$ of $\mathbf{\text{0.400 M}}$ sodium acetate. (See Appendix C.) (a) What is the pH? (b) How many grams of $\mathbf{\text{KOH}}$ must be added to $\mathbf{\text{0.500 L}}$ of the buffer to change the pH by $\mathbf{\text{0.15}}$ units?

Step-by-Step Solution

Answer

1. The value of  $\text{pH = 4.81}$.
2. The mass required is  $\text{0.67 g KOH}$.

Step 1: Buffer solution

A buffer is a solution that can withstand pH changes when acidic or basic components are added. It can neutralize little amounts of additional acid or base, allowing the pH of the solution to remain relatively constant.

Step 2: Explanation

(a)

First, write the HCL and ${\text{CH}}_{\text{3}}\text{COONa}$ dissociation.

$$\begin{gathered} {\text{ HCl + H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{ + Cl}}^{\text{ - }} \\ {\text{CH}}_{\text{3}}\text{COONa}\to {\text{CH}}_{\text{3}}{\text{COO}}^{\text{ - }}{\text{ + Na}}^{\text{ + }} \end{gathered}$$

 

Calculate the amount of mol of hydronium ion and acetate in the equation.

 $$\begin{gathered} \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{\mathrm{mol}}{\mathrm{L}} \\ \mathrm{mol} {\mathrm{H}}_3{\mathrm{O}}^{+}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\times \mathrm{L} \\ =(0.452\mathrm{M})\left(204 \mathrm{mL}\times \frac{1 \mathrm{L}}{1000 \mathrm{mL}}\right) \\ \mathrm{mol} {\mathrm{H}}_3{\mathrm{O}}^{+}=0.0922 \mathrm{mol} \\ \left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]=\frac{\mathrm{mol}}{\mathrm{L}} \\ \mathrm{mol} {\mathrm{CH}}_3{\mathrm{COO}}^{-}=\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]\times \mathrm{L} \\ =(0.400 \mathrm{M})(0.500 \mathrm{L}) \\ \mathrm{mol} {\mathrm{CH}}_3{\mathrm{COO}}^{-}=0.200 \mathrm{mol} \end{gathered}$$

After that, acetate reacts with the hydronium ion to produce acetic acid. The reaction will complete because the hydronium ion is a strong acid. The entire amount of strong acid will be consumed.

${\mathrm{H}}_3{\mathrm{O}}^{+}+{\mathrm{CH}}_3{\mathrm{COO}}^{-}\to {\mathrm{CH}}_3\mathrm{COOH}+{\mathrm{H}}_2\mathrm{O}$

 

The reaction also created  ${\text{CH}}_{\text{3}}\text{COOH}$, as seen. Since all ${\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}$ has been consumed, the concentration of [${\text{CH}}_{\text{3}}{\text{COO}}^{\text{ - }}$] has decreased by $\text{0.0922 M}$, and the concentration of [${\text{CH}}_{\text{3}}\text{COOH}$] has decreased by $\text{0.0922 M}$.

Step 3: Evaluating the pH

Now, using the ${\text{K}}_{\text{a}}$ from Appendix C, find the ${\text{pK}}_{\text{a}}$ of acetic acid.

 $$\begin{gathered} {\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}} \\ =-\log \left(1.8\times {10}^{-5}\right) \\ {\mathrm{pK}}_{\mathrm{a}}=4.74 \end{gathered}$$

The concentrations of  ${\text{CH}}_{\text{3}}\text{COOH}$ and  ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{ - }}$ are then calculated.

Total volume is,

$0.500 \text{L}+\left(204 \text{mL}\times \frac{1 \text{L}}{1000\text{mL}}\right)=0.704 \text{L}$

$$\begin{gathered} \left[{\text{CH}}_3\mathrm{COOH}\right]=\frac{\mathrm{mol}}{\mathrm{L}} \\ =\frac{0.0922\text{ mol}}{0.704 \text{L}} \\ =0.131 \text{M} \\ \left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]=\frac{\text{mol}}{\text{L}} \\ =\frac{0.200-0.0922 \text{mol}}{0.704\text{ L}} \\ =0.153 \text{M} \end{gathered}$$

 

Calculate the pH now.

 $$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]} \\ =4.74+\log \frac{0.153}{0.131} \\ \mathrm{pH}=4.81 \end{gathered}$$

Therefore,  $\text{pH = 4.81}$.

Step 4: Explanation

(b)

The pH will rise as a result of the addition of a strong base. As a result, we'll need to raise the pH by $0.15$ units.

 $\text{pH +} 0.15 {\text{= pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{ - }}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}$

It's important to remember that the additional base will react with ${\text{CH}}_{\text{3}}\text{COOH}$. $\text{KOH}$ will be the first to dissociate.

 $$\begin{gathered} \text{ KOH}\to {\text{K}}^{\text{ + }}{\text{ + OH}}^{\text{ - }} \\ {\text{CH}}_{\text{3}}{\text{COOH + OH}}^{\text{ - }}\to {\text{CH}}_{\text{3}}{\text{COO}}^{\text{ - }}{\text{ + H}}_{\text{2}}\text{O} \end{gathered}$$

The reaction also created  ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{ - }}$, as seen. Since all of the ${\text{OH}}^{\text{ - }}$ has been consumed, the concentration of ${\text{CH}}_{\text{3}}\text{COOH}$ will decrease by an unknown amount, which we shall refer to as x. The equal amount of ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{ - }}$ will be created in addition.

As a result, our working equation will be...

${\text{pH + 0.15 = pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{\text{ - }}\right]\text{ + x}}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]\text{ - x}}$

 

Now that we know all of the variables' values in the equation for x, we can solve for x.

$$\begin{gathered} \mathrm{pH}+0.15={\text{pK}}_a+\log \frac{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]+\mathrm{x}}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]-\mathrm{x}} \\ \text{log}\frac{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]+\mathrm{x}}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]-\mathrm{x}}=\text{pH}+0.15-{\text{pK}}_a \\ \frac{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]+\mathrm{x}}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]-\mathrm{x}}={10}^{\text{pH}+0.15-{\text{pK}}_a} \end{gathered}$$ 

Step 5: Evaluating the mass

Replace the values first. Then manipulate with the equation.

$$\begin{gathered} \frac{0.153+x}{0.131-x}={10}^{4.81+0.15-4.74} \\ \frac{0.153+x}{0.131-x}=1.660 \\ 0.153+x=1.660(0.131-x) \\ x+1.660x=0.217-0.153 \\ 2.660x=0.064 \\ x=0.024 \text{M} \end{gathered}$$

 

As a result, $\text{0.024 M}$ KOH must be added. Calculate the required mass.

 $$\begin{gathered} m=0.024\frac{\text{mol KOH}}{\text{L}}\times 0.500 \text{L}\times \frac{56.10\text{ g KOH}}{1 \text{mol KOH}} \\ =0.67 \text{g KOH} \end{gathered}$$

Therefore, the mass required is $\text{0.67 g KOH}$.