Question

1 ml of ${10}^{-5}$ M HCl solution is diluted to 1000 ml. Then the pH of the diluted solution is : $(K_w={10}^{-14}andlog(10.5)=1.02$) (Consider only two points after decimal.)

Answer 1

When 1 ml of ${10}^{-5}$ M HCl is diluted to 1000 ml, then the concentration of HCl will be ${10}^{-8}$ M.
Now the pH of HCl would be 8 which comes in the basic range of pH. But it isn't true.
So, it should be noted that in this case while calculating pH we have to consider the $H^{+}$ ions from the dissociation of water in addition to the $H^{+}$ ions given by HCl.
Here the dissociation of water is surpressed due to the commom-ion effect.
$H_{2}O\rightleftharpoons H^{+}+O{H}^{-}$
x x
where x is $\left[H{]}^{+}or\right[OH{]}^{-}$
$\therefore Totalconcentrationof{H}^{+}ions=(x+{10}^{-8})$
$(\because [H{]}^{+}=\left[HCl\right]={10}^{-8})$
Now $K_w=\left[H^{+}\right][OH{]}^{-}{10}^{-14}=(x+{10}^{-8})x\Rightarrow x=9.5\times {10}^{-8}$

$\therefore Total[H{]}^{+}=(9.5\times {10}^{-8}+{10}^{-8})$
Thus $pH=-log[H{]}^{+}=-log(9.5\times {10}^{-8}+{10}^{-8})=6.98$