$PH=$Pka$+log\frac{A^{-}}{HA},C_1{V}_1=C_2{V}_2$

$2HCl=0\mathrm{.}002ml,PH=6\mathrm{.}9,$pKa$2=7\mathrm{.}2$

$0\mathrm{.}002ml10=100C_2$

$C_2=0\mathrm{.}002\frac{10}{100}=0\mathrm{.}0002M$

$100ml$ of $50mM=0\mathrm{.}025$

$A^{-}=0\mathrm{.}025-0\mathrm{.}0001=0\mathrm{.}0249$

$[HA]=0\mathrm{.}025+0\mathrm{.}0001=0\mathrm{.}0251$

$PH=7\mathrm{.}2+log(\frac{0\mathrm{.}0249}{0\mathrm{.}0751})=7\mathrm{.}196$

$HC=0\mathrm{.}0002M$

$PH=-log(0\mathrm{.}0002)=3\mathrm{.}7$

If the concentration of the buffer were $250$mM$,$ what would the change in pH be$?$