50ml of a weak monoprotic acid was titrated with 0.1M solution of NaOH. If pH of the solution after adding 50 ml and 75 ml of base is 4.699 and 5 then calculate the value x+y where: x=pKa of the weak acid; y=pH of the solution when 7.5 millimoles of HCl are added in the solution at the equivalence point without changing volume.[Given: $\log 2=0.301]$

$\because$ after adding of NaOH to weak monoprotic acid medium i.e NaOH added is quite lesser After addition of 50 ml NaoH 0.1M
$\begin{array}{l}HA+NaOH\to NaA+H_{2}O\\ (50\text{\hspace{0.17em}ml,}\times \text{M})(50\text{ml},0.1)\end{array}$

Left moles in solution (50x-5)

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}\\ 4.699=p{K}_a+\log \left(\frac{5}{50x-5}\right)\mathrm{....}\left(i\right)\end{array}$

After adding of 75 ml NaOH 0.1M
Left millimoles of HA=50x-7.5
Millimoles of NaA formed=7.5
$\begin{array}{l}pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}\\ 5=p{K}_a+\log \left(\frac{7.5}{50x-75}\right)\mathrm{...}\left(ii\right)\end{array}$

On subtracting (ii) – (i), $0.3010=\log \left(\frac{7.5}{50x-7.5}\frac{(50x-5)}{5}\right)=\log 2$

On subtracting x=0.3

$NaA+ HCl\to NaCl+HA$

15 Millimioles 7.5 Millimoles

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[A^{-}\right]}{\left[HA\right]}=p{K}_a+\log \frac{7.5}{7.5}\\ pH=p{K}_a=5\Rightarrow cd=05\end{array}$