Question

A buffer that contains 1.05 M B and 0.750 M BH⁺ has a pH of 9.50. What is the pH after 0.0050 mol of HCl is added to 0.500 L of this solution?

Answer 1

$First, solve for the p K_a of the base by manipulating the Henderson-Hasslebalch equation.$

$\begin{aligned}p H &-p K_a+\log \frac{[B]}{\left[B H^{+} \mid\right.} \\p K_a &-p H-\log \frac{[B]}{\left[B H^{+} \mid\right.} \\&-9.50-\log \frac{[1.05]}{[0.750]} \\p K_a &-9.35\end{aligned}$

$Now solve for the Molarity of the added \mathrm{HCl} . M-\frac{m o l}{L}-\frac{0.0050 \mathrm{~mol}}{0.500 \mathrm{~L}}-0.01 \mathrm{M} H C l will dissociate in water.$

$\mathrm{HCl}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{Cl}^{-} Then as a strong acid, all \mathrm{H}_3 \mathrm{O}^{+} will react with B . \mathrm{B}+\mathrm{H}_3 \mathrm{O}^{+} \rightarrow \mathrm{BH}^{+}+\mathrm{H}_2 \mathrm{O}$

$As observed, the reaction also produced \mathrm{BH}^{+} . Since all \mathrm{H}_3 \mathrm{O}^{+} were consumed, \mathrm{B} will decrease in concentration by 0.01 M and B H^{+} will increase in concentration by 0.01 M .|B|-1.05-0.01-1.04 M \left[B H^{+}\right]-0.750+0.01-0.760 M Now solve for the new p H using the p K_a and the new values of [B] and \left[B H^{+}\right] .Now solve for the new p H using the p K_a and the new values of [B] and \left[B H^{+}\right] .$

$\begin{aligned}p H &-p K_a+\log \frac{|B|}{\left[B H^{+} \mid\right.} \\&-9.35+\log \frac{1.04}{0.760} \\p H-9.49\end{aligned}$

What is a buffer?

Buffer is a solution which resists change of pH upon addition of a small amount of acid or base. Many chemical reactions are carried out at a constant pH. Numerous pH regulating systems in nature employ buffering. For instance, the pH of blood is controlled by the bicarbonate buffering system, and bicarbonate also serves as a buffer in the ocean. There are two types of buffer: acidic and basic buffer.

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