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100 mL of each pH = 3 and pH = 5 solutio...

100 mL of each pH = 3 and pH = 5 solutions of a strong acid are mixed. What is the resultant pH ?

Text Solution

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pH of first solution `= 3, [H^(+) ]=10^(-3) =N_1`
pH of second solution `= 5, [H^(+) ]=10^(-5) = N_2`
Final proton concentration is given by the equation:`[H^+]=(V_1 N_1+V_2 N_2)/(V_1 +V_2)`
Substituting the values` [H^(+)] =(100 xx 10^(-3) xx 10^(-5))/( 100+100 ) = 5.05 xx 10^(-4)` M
pH of the resultant mixture `= -log (5.05 xx 10^(-4) ) = 4 - log 5.05 = 3.3`
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