To a 50 ml of `0.1` M HCl solution , 10 ml of `0.1` M NaOH is added and the resulting solution is diluted to 100 ml. What is change in pH of the HCl solution ?

Options

A. `4.398`
B. `0.398`
C. `0.1 M`
D. None of these

Answer

2

Explanation

Before adding HCl solution
`pH = 1 [HCl] = [H_(3)O]^(+) = 10^(-1) M]`
`n_("HCl") "(initially)" = MV = 0.1 M xx 0.05 L = 5 xx 10^(-3) ` mol
`n_("NaOH") " added " = MV = 0.1 M xx 0.01L = 1 xx 10^(-3) ` mol
`{:(" "HCl,+,NaOH to ,NaCl,+,H_(2)O),(t=0 " "5 xx 10^(-3) mol,,1xx10^(-3) mol,0,,-),(t=t_("final") 4 xx10^(-3) mol,," "0,1xx10^(-3)mol,,-),(V_("final") = 100 ml = 0.1L,,,,,):}`
`[HCl] = n/V = (4 xx 10^(-3) mol)/(0.1L) = 4 xx 10^(-2) M`
`pH = - log [H_(3)O^(+)]= 2 - log4 = 2 - 2 log 2 = 2 - 2 xx 0.301 = 2 - 0.602 = 1.398`
Increase in `pH = (1.398 -1) = 0.398`.