$\ce{NH3}$ solution of $\pu{0.1 mol dm-3}$ is being added to a $\pu{25.0 cm3}$ of $\pu{0.1 mol dm-3}$ $\ce{HCl}$ solution. Calculate the pH of the solution when volume of added $\ce{NH3}$ solution is $\pu{25.0 cm3}$ and $\pu{26.0 cm3}$. The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is ${9.25}$.
So I started with the equation:
$$\ce{HCl + NH3 <=> NH4Cl}\tag{1}$$
EDIT #1:
As the initial concentration of both base and acid are same, I calculated how much of base is left in the solution:
$V_\text{miscon} = \pu{26.0 cm3 - 25.0 cm3 = 1.0 cm3}$
That's how I tried to calculate pH of the solution when $\pu{26.0 cm3}$ of $\ce{NH3}$ is added:
$\ce{H+}$ concentration: $$[\ce{H+}]_\text{miscon} = [\ce{H+}]_\text{initial} \frac{V_\text{initial}}{V_\text{miscon}}\tag{2} $$ $$= \pu{(0.1 mol dm-3} \times \pu{0.025 dm3)/0.001 dm3} = \pu{2.5 mol dm-3}$$
But, in this way, the results I am getting are completely different from marking, and even negative pH values.
But, when I tried to find $\ce{H+}$ concentration (to calculate pH) of the solution when $\pu{25.0 cm3}$ of $\ce{NH3}$ is added, I found out that initial $\ce{HCl}$ volume is completely reacted. Only ammonium chloride is left in the solution.
The marking only gives the answers, $\ce{pH = 5.28}$ and $\ce{pH = 7.85}$.
How can I calculate pH of this solution in above two conditions separately?
I can't figure it out, I am stuck in the above equation. Please give me some hints to start solving. Thanks in advance.
EDIT #2:
I finally figured it out (Adding $\pu{25.0 cm3}$ of $\ce{NH3}$).
$$\ce{[NH4+(aq)]}=n(\ce{NH4+})/V_{\mathrm{\ce{total}}}$$ $$=(\pu{0.1 mol dm-3}\cdot\pu{0.025 dm3})/\pu{0.05 dm3}$$ $$=\pu{0.0025 mol}/\pu{0.05 dm3}$$ $$\ \ \ =5.0\times 10^{-2}\ \pu{mol dm-3}$$
$\ce{NH4+}$ reacts with water to produce $\ce{H3O+}$ ions.
$$\ce{NH4+(aq) + H2O(l) <=> H3O+(aq) + NH3(aq)}$$
Using acid dissociation constant for $\ce{NH4+}$:
$$\ce{K_\mathrm{a} = [H3O+(aq)] [NH3(aq)] / [NH4+(aq)]}$$ $$\ce{[H3O+(aq)] = \pu{( K_\mathrm{a} \cdot [\ce{NH4+(aq)}] )0.5}}$$ $$\ce{= \pu{( 5.62\times 10^{-10}\pu{mol dm-3} \cdot 5.0\times 10^{-2}\pu{mol dm-3} )0.5}}$$ $$\ce{= \pu{( 2.81\times 10^{-11}\pu{mol2 dm-6} )0.5}}$$ $$\ \ \ =5.301\times 10^{-6}\ \pu{mol dm-3}$$
$$ \text{pH} = -(\log 5.301 - 6\log 10) = -(0.7244 - 6) = 5.28 $$