Water is added to $1.0 \ mL$ of $0.1 \ M$ $HCl$ solution to make the total volume $50 \ mL$. Calculate the $pH$ change of the solution.
(N/A) Initial state: $V_1 = 1.0 \ mL$,$M_1 = 0.1 \ M$
Final state: $V_2 = 50 \ mL$,$M_2 = ?$
Using the dilution equation: $M_1 V_1 = M_2 V_2$
$0.1 \ M \times 1.0 \ mL = M_2 \times 50 \ mL$
$M_2 = \frac{0.1}{50} = 0.002 \ M = 2 \times 10^{-3} \ M$
Initial $pH = -\log(0.1) = 1$
Final $pH = -\log(2 \times 10^{-3}) = 3 - \log(2) = 3 - 0.301 = 2.699$
Change in $pH = 2.699 - 1 = 1.699$
Final state: $V_2 = 50 \ mL$,$M_2 = ?$
Using the dilution equation: $M_1 V_1 = M_2 V_2$
$0.1 \ M \times 1.0 \ mL = M_2 \times 50 \ mL$
$M_2 = \frac{0.1}{50} = 0.002 \ M = 2 \times 10^{-3} \ M$
Initial $pH = -\log(0.1) = 1$
Final $pH = -\log(2 \times 10^{-3}) = 3 - \log(2) = 3 - 0.301 = 2.699$
Change in $pH = 2.699 - 1 = 1.699$
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