Q19.32P - Chemistry: Molecular Nature Of Matter And Change

Question

A buffer is prepared by mixing $\mathbf{\text{50.0 mL}}$ of $\mathbf{\text{0.050 M}}$ sodium bicarbonate and $\mathbf{\text{10.7 mL}}$ of  $\mathbf{\text{0.10 M NaOH}}$. (See Appendix C.) (a) What is the pH? (b) How many grams of HCL must be added to $\mathbf{\text{25.0 mL}}$ of the buffer to change the pH by $\mathbf{\text{0.07}}$ units?

Step-by-Step Solution

Answer

1. The value of $\text{pH = 10.15}$.
2. The mass required is $\text{0.00143 g HCl}$.

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 ${\text{NaHCO}}_{\text{3}}$ and $\text{NaOH}$ dissociation.

 $$\begin{gathered} {\text{NaHCO}}_{\text{3}}\to {\text{Na}}^{\text{ + }}{\text{ + HCO}}_{\text{3}}^{\text{ - }} \\ \text{NaOH}\to {\text{OH}}^{\text{ - }}{\text{ + Na}}^{\text{ + }} \end{gathered}$$

Calculate the amount of mol of bicarbonate ion and hydroxide ion.

 $$\begin{gathered} \left[{\mathrm{HCO}}_3^{-}\right]=\frac{\mathrm{mol}}{\mathrm{L}} \\ {\text{mol HCO}}_3^{-}=\left[{\mathrm{HCO}}_3^{-}\right]\times \mathrm{L} \\ =(0.050\text{ M})\left(50.0\text{ mL}\times \frac{1\text{ L}}{1000\text{ mL}}\right) \\ {\text{mol HCO}}_3^{-}=0.0025 \text{mol} \\ \left[{\mathrm{OH}}^{-}\right]=\frac{\mathrm{mol}}{\mathrm{L}} \\ {\text{mol OH}}^{-}=\left[{\mathrm{OH}}^{-}\right]\times \mathrm{L} \\ =(0.1 \text{M})\times \left(10.0 \text{mL}\times \frac{1 \text{L}}{1000 \text{mL}}\right) \\ {\text{mol OH}}^{-}=0.001 \text{mol} \end{gathered}$$

After that, bicarbonate reacts with the hydroxide ion to produce carbonic acid. The reaction will complete because the hydroxide ion is a strong base. The entire amount of strong base will be consumed.

 ${\text{HCO}}_{\text{3}}^{\text{ - }}{\text{ + OH}}^{\text{ - }}\to {\text{CO}}_{\text{3}}^{\text{2 - }}{\text{ + H}}_{\text{2}}\text{O}$

The reaction also created ${\text{CO}}_{\text{3}}^{\text{2 - }}$, as seen. Since all ${\text{OH}}^{\text{ - }}$ has been consumed, the concentration of $\left[{\text{HCO}}_{\text{3}}^{\text{ - }}\right]$ has decreased by $\text{0.001 M}$, and the concentration of [${\text{CO}}_{\text{3}}^{\text{2 - }}$] is $\text{0.001 M}$.

Step 3: Evaluating the pH

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

$$\begin{gathered} {\mathrm{pK}}_{\mathrm{a}}=-{\mathrm{logK}}_{\mathrm{a}} \\ =-\log \left(4.7\times {10}^{-11}\right) \\ {\mathrm{pK}}_{\mathrm{a}}=10.33 \end{gathered}$$

 

The concentrations of ${\text{HCO}}_{\text{3}}^{\text{ - }}$ and ${\text{CO}}_{\text{3}}^{\text{2 - }}$ are then calculated.

Total volume is,

$\left(50.0\text{ mL}\times \frac{1\text{ L}}{1000\text{ mL}}\right)+\left(10.7\text{ mL}\times \frac{1\text{ L}}{1000\text{ mL}}\right)=0.0607\text{ L}$ 

 $$\begin{gathered} \left[{\mathrm{CO}}_3^{2-}\right]=\frac{\mathrm{mol}}{\mathrm{L}} \\ =\frac{0.001 \text{mol}}{0.0607 \text{L}} \\ =0.0165\text{ M} \\ \left[{\mathrm{HCO}}_3^{-}\right]=\frac{\mathrm{mol}}{\mathrm{L}} \\ =\frac{0.0025-0.001\text{ mol}}{0.0607 \text{L}} \\ =0.0247\text{ M} \end{gathered}$$

Calculate the pH now.

 $$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\log \frac{\left[{\mathrm{CO}}_3^{2-}\right]}{\left[{\mathrm{HCO}}_3^{-}\right]} \\ =10.33+\log \frac{0.0165}{0.0247} \\ \mathrm{pH}=10.15 \end{gathered}$$

Therefore, $\text{pH = 10.15}$.

Step 4: Explanation

(b)

The pH will decrease as a result of the addition of a strong acid. As a result, we'll need to subtract the   by $\text{0.07}$ units.

${\text{pH - 0.07 = pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{CO}}_{\text{3}}^{\text{2 - }}\right]}{\left[{\text{HCO}}_{\text{3}}^{\text{ - }}\right]}$

 

It's important to remember that the additional acid will react with ${\text{CO}}_{\text{3}}^{\text{2 - }}$. $\text{HCl}$ will be the first to dissociate.

 $$\begin{gathered} {\text{ HCl + H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{ + Cl}}^{\text{ - }} \\ {\text{CO}}_{\text{3}}^{\text{2 - }}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{ + }}\to {\text{HCO}}_{\text{3}}^{\text{ - }}{\text{ + H}}_{\text{2}}\text{O} \end{gathered}$$

The reaction also created ${\text{HCO}}_{\text{3}}^{\text{ - }}$, as seen. Since all of the ${\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}$ has been consumed, the concentration of ${\text{CO}}_{\text{3}}^{\text{2 - }}$ will decrease by an unknown amount, which we shall refer to as x. The equal amount of  ${\text{HCO}}_{\text{3}}^{\text{ - }}$ will be created in addition.

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

 ${\text{pH - 0.07 = pK}}_{\text{a}}\text{ + log}\frac{\left[{\text{CO}}_{\text{3}}^{\text{2 - }}\right]\text{ - x}}{\left[{\text{HCO}}_{\text{3}}^{\text{ - }}\right]\text{ + x}}$

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

 $$\begin{gathered} \mathrm{pH}-0.07={\text{pK}}_a+\log \frac{\left[{\mathrm{CO}}_3^{2-}\right]-\mathrm{x}}{\left[{\mathrm{HCO}}_3^{-}\right]+\mathrm{x}} \\ \log \frac{\left[{\mathrm{CO}}_3^{2-}\right]-\mathrm{x}}{\left[{\mathrm{HCO}}_3^{-}\right]+\mathrm{x}}=\text{pH}-0.07-{\text{pK}}_a \\ \frac{\left[{\mathrm{CO}}_3^{2-}\right]-\mathrm{x}}{\left[{\mathrm{HCO}}_3^{-}\right]+\mathrm{x}}={10}^{\text{pH}-0.07-{\text{pK}}_a} \end{gathered}$$

Step 5: Evaluating the mass

Replace the values first. Then manipulate with the equation.

$$\begin{gathered} \frac{[0.0165]-x}{[0.0247]+x}={10}^{10.15-0.07-10.33} \\ \frac{[0.0165]-x}{[0.0247]+x}=0.562 \\ 0.0165-x=0562(0.0247+x) \\ 0.562x+x=0.0165-0.0139 \\ 1.562x=0.0026 \\ x=0.00166\text{ M} \end{gathered}$$ 

As a result, $\text{0.00166 M}$ $\text{HCL}$ must be added. Calculate the required mass.

 $$\begin{gathered} m=0.00166\frac{\text{mol HCl}}{\text{L}}\times 25.0 mL\times \frac{1\text{ L}}{1000\text{ mL}}\times \frac{36.46\text{ g HCl}}{1 \text{mol HCl}} \\ =0.00143\text{ g HCl} \end{gathered}$$

Therefore, the mass required is  $\text{0.00143 g HCl}$.