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# A solution is obtained by mixing two solutions of same electrolyte with pH = 5 and pH = 3 respectively. The resulting solution has pH

(a) 2.2

(b) 4.0

(c) 8.0

(d) 3.3

Toolbox:
• $\rho H = - log [H^+]$
If $\rho H = 5 H^+ = 10^{-\rho H} = 10^{-5}$
If $\rho H = 3 H^+ =10^{-3}$
Total  $[H^+] = \frac{10^{-5} + 10^{-3}}{2}$ = $\frac{.01 \times 10^{-3} + 1 \times 10^{-3}}{2}$ = $\frac{(1 + 0.01) \times 10^{-3}}{2}$
$\therefore [H^+] = \frac{1.01 \times 10^{-3}}{2} = 0.505 \times 10^{-3}$
$\rho H = - log (0.505 \times 10^{-3}) = 3.3$
edited Mar 25, 2014