- #1

- 254

- 0

100 mL 0,102 M HCL and 100 mL 0,0780 M NaOH.

To calculate the pH in the this solution first I must calculate the number of moles [itex]n_{[H_3O^+]}[/itex].

[itex]n_{[H_3 O^{+}]} = 0,100 L \cdot 0,0102 \ mol/L + 0,100 L \cdot 0,0780 mol/L = 0,018 mol[/itex]

This means that [itex][H_3 O^{+}] = \frac{0,018 mol}{0,200 L} = 0,09 mol/L[/itex]

pH for the solution is then [itex]pH = \textrm{-log}(0,09) = 1,05[/itex]

I would appriciate if somebody would look at my calculation and then tell me if its accurate ??

Many thanks in advance.

Sincerely

Fred