What would be the final pH if I add 25 ml of water to 25 ml of a 0.1N HCl?

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What would be the final pH if I add 25 ml of water to 25 ml of a 0.1N HCl?

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Answer 1 · 2018-02-01T19:52:09

$p H_{final} = 1.30$ $pH_"final"=1.30$

Explanation:

By definition, $p H = - {log}_{10} [H_{3} O^{+}]$ $pH=-log_10[H_3O^+]$

Initially, we gots $0.1 \cdot m o l \cdot L^{- 1}$ $0.1*mol*L^-1$ $H C l$ $HCl$ ...

...and $- {log}_{10} (0.1) = - {log}_{10} (10^{- 1}) = - (- 1) = 1$ $-log_10(0.1)=-log_10(10^-1)=-(-1)=1$ ...

We dilute the solution by half....

$[H_{3} O^{+}] = \frac{25 \times 10^{- 3} \cdot L \times 0.10 \cdot m o l \cdot L^{- 1}}{50 \times 10^{- 3} \cdot L} = 0.05 \cdot m o l \cdot L^{- 1}$ $[H_3O^+]=(25xx10^-3*Lxx0.10*mol*L^-1)/(50xx10^-3*L)=0.05*mol*L^-1$ ...

$p H = - {log}_{10} (0.05) = 1.30$ $pH=-log_10(0.05)=1.30$

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What would be the final pH if I add 25 ml of water to 25 ml of a 0.1N HCl?

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