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What volume of 0.01mol NaOH must be added to 5ml of 0.005M of HCl if the resulting solution have a pH of 10.796?

I will be glad if you help me with the analysis.
Thanks!

Uricas:
What volume of 0.01mol NaOH must be added to 5ml of 0.005M of HCl if the resulting solution have a pH of 10.796?

I will be glad if you help me with the analysis.
Thanks!

Your V1 is unknown, V2 is 5, m1 is 0.01, m2 is 0.005 with the pH of d result as 10.796, you calculate the no of moles for each reactant, n = CV
now when you react both HCl and NaOH, you have salt and water and stoichiometry as 1:1
You'll get 0.001V/0.025, equate it to the pH
and from there you get your volume
pls check other sources, I'm not too sure, I'll read up on it and see if I'm correct.

opeaceo:
Your V1 is unknown, V2 is 5, m1 is 0.01, m2 is 0.005 with the pH of d result as 10.796, you calculate the no of moles for each reactant, n = CV
now when you react both HCl and NaOH, you have salt and water and stoichiometry as 1:1
You'll get 0.001V/0.025, equate it to the pH
and from there you get your volume
pls check other sources, I'm not too sure, I'll read up on it and see if I'm correct.

thanks, but what you referred to as m1 is actually amount in mole not malarity.

Uricas:
What volume of 0.01mol NaOH must be added to 5ml of 0.005M of HCl if the resulting solution have a pH of 10.796?

I will be glad if you help me with the analysis.
Thanks!

using CaVa/CbVb=Na/Nb
Ca=0.005m
Va=5ml
Cb=?
Vb=?
Na=1
Nb=1
Since pH=10.796
Den poH=14-10.796
=3.204.
BT poH=-log[Cb]
3.204=-log[Cb]
Cb=6.25*10^-4M
Substituting dat to d equation above with the given values
Vb=40ml.