Calculate the $pH$ of a mixed solution containing $10 \ mL$ of $0.2 \ M$ $Ca(OH)_2$ and $25 \ mL$ of $0.1 \ M$ $HCl$.

$100 \ cm^3$ of $0.1 \ N \ HCl$ solution is mixed with $100 \ cm^3$ of $0.2 \ N \ NaOH$ solution. The resulting solution is

At $25^{\circ} C$,the ionic product of water is $10^{-14}$. The free energy change for the self-ionization of water in $kCal \ mol^{-1}$ is close to

At $27^{\circ}C$,$100 \ mL$ of $0.4 \ M$ $HCl$ is mixed with $100 \ mL$ of $0.5 \ M$ $NaOH$ solution. To the resultant solution,$800 \ mL$ of distilled water is added. What is the $pH$ of the final solution?

Calculate the $pH$ of the resultant mixtures:
$(a)$ $10 \, mL$ of $0.2 \, M \, Ca(OH)_{2} + 25 \, mL$ of $0.1 \, M \, HCl$
$(b)$ $10 \, mL$ of $0.01 \, M \, H_{2}SO_{4} + 10 \, mL$ of $0.01 \, M \, Ca(OH)_{2}$
$(c)$ $10 \, mL$ of $0.1 \, M \, H_{2}SO_{4} + 10 \, mL$ of $0.1 \, M \, KOH$

$A$ solution of $0.1 \ M$ weak base $(B)$ is titrated with $0.1 \ M$ of a strong acid $(HA)$. The variation of $pH$ of the solution with the volume of $HA$ added is shown in the figure below. What is the $pK_{b}$ of the base? The neutralization reaction is given by $B + HA \rightarrow BH^{+} + A^{-}$.