50 mL of 0.1 M HCl and 50 mL of 2.0 M NaOH are mixed. Calculate the pH of the resulting solution. Consider log 2=0.30A. 1.3B. 4.2C. 12.7D. 11.7
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`4.398` `0.398``0.1 M` None of these
Answer : 2
Solution : Before adding HCl solution <br> `pH = 1 [HCl] = [H_(3)O]^(+) = 10^(-1) M]` <br> `n_("HCl") "(initially)" = MV = 0.1 M xx 0.05 L = 5 xx 10^(-3) ` mol <br> `n_("NaOH") " added " = MV = 0.1 M xx 0.01L = 1 xx 10^(-3) ` mol <br> `{:(" "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,,,,,):}`<br> `[HCl] = n/V = (4 xx 10^(-3) mol)/(0.1L) = 4 xx 10^(-2) M` <br> `pH = - log [H_(3)O^(+)]= 2 - log4 = 2 - 2 log 2 = 2 - 2 xx 0.301 = 2 - 0.602 = 1.398` <br> Increase in `pH = (1.398 -1) = 0.398`.