Mathman23
Nov23-04, 01:10 PM
I have calculated the PH-value for this following solution:
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 n_{[H_3O^+]}.
n_{[H_3 O^{+}]} = 0,100 L \cdot 0,0102 \ mol/L + 0,100 L \cdot 0,0780 mol/L = 0,018 mol
This means that [H_3 O^{+}] = \frac{0,018 mol}{0,200 L} = 0,09 mol/L
pH for the solution is then pH = \textrm{-log}(0,09) = 1,05
I would appriciate if somebody would look at my calculation and then tell me if its accurate ??
Many thanks in advance.
Sincerely
Fred
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 n_{[H_3O^+]}.
n_{[H_3 O^{+}]} = 0,100 L \cdot 0,0102 \ mol/L + 0,100 L \cdot 0,0780 mol/L = 0,018 mol
This means that [H_3 O^{+}] = \frac{0,018 mol}{0,200 L} = 0,09 mol/L
pH for the solution is then pH = \textrm{-log}(0,09) = 1,05
I would appriciate if somebody would look at my calculation and then tell me if its accurate ??
Many thanks in advance.
Sincerely
Fred