[Lectos]

I have the following problem. I have solved it making estequiometric considerations. I am trying to solve it using gram equivalents but I cannot get to the correct answer. Some guidance will be apreciated.

The problem says as follows
Calculate the pH of a solution that was prepared by adding 50 ml of NaOH 0.10M to 100 ml of H₃PO₄ 0.10M

And I did the following

What I did was first get the grams of each compund:
g of NaOH = (0.1M)(40 g/mol)(0.05 L) = 0.2g
g of H₃PO₄ = (0.1M)(98 g/mol)(0.1 L) = 0.98

With this I calculated the gram equivalents

eq of NaOH = 5x10^-3
eq of H₃PO₄ = 0.03

After this I substracted the equivalents of acid form the base

0.03-5x10^-3 = 0.025

And because this will be a buffer solution

eq. of H₃PO₄ = 0.025
eq. of H₂PO₄ = 0.005

And...

[H+] = Ka₁ [acid]/[salt] = 0.0375
pH = 1.42

The correct answer sould be 2.28

I know I did something wrong but I cannot see what it is.

[Borek]

What I did was first get the grams of each compund:
g of NaOH = (0.1M)(40 g/mol)(0.05 L) = 0.2g
g of H₃PO₄ = (0.1M)(98 g/mol)(0.1 L) = 0.98

Correct, but not needed.

With this I calculated the gram equivalents

eq of NaOH = 5x10^-3
eq of H₃PO₄ = 0.03

These can be calculated just from number of moles, that is - concentration times volume.

0.03-5x10^-3 = 0.025

And because this will be a buffer solution

eq. of H₃PO₄ = 0.025

No, it it is not 0.025. You are left with 0.01+0.01+0.005 equivalents of different acids.