Answer :
[tex]$ First, solve for the $p K_a$ of the base by manipulating the Henderson-Hasslebalch equation.$[/tex]
[tex]$\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}$[/tex]
[tex]$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.$[/tex]
[tex]$\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 $B$.$\mathrm{B}+\mathrm{H}_3 \mathrm{O}^{+} \rightarrow \mathrm{BH}^{+}+\mathrm{H}_2 \mathrm{O}$[/tex]
[tex]$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 $\left[B H^{+}\right]$.$[/tex]
[tex]$\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}$[/tex]
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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