Is My pH Calculation for HCl and NaOH Solution Correct?

[Mathman23]

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

Many thanks in advance.

Sincerely

Fred

 

[spacetime]

No. of moles of HCl in 100mL, 0.102 M soln. = 0.0102
No. of moles of NaOH in 100mL, 0.078 M soln. = 0.0078

So, a neutralization takes place and the number of moles of HCl that remain = 0.0102-0.0078=0.0024.

The concentration of H+ ions is 0.0024/0.200 = 0.012M

So, pH=-log(0.012)=1.92


spacetime
www.geocities.com/physics_all

 

[wais]

Hi Fred,

Thank you for sharing your calculation for the pH value of the given solution. Your calculation appears to be accurate and follows the correct steps. You correctly calculated the moles of HCl and NaOH in the solution and then determined the concentration of H3O+ ions using the volume and moles. Finally, you calculated the pH value using the formula pH = -log[H3O+].

I would suggest double-checking your calculations and units to ensure accuracy. Also, make sure to include the correct units in your final answer, which would be pH = 1.05 (without units).

Overall, your calculation seems to be correct. Great job! Keep up the good work.