pH equation (1 Viewer)

[kyokaku92]

I need some help with the following eqn:

120mls of 0.2 mol/L of H2SO4 was added to 65mls of 0.08 mol/L of KOH. Calculate pH.

Cheers

 

[gurmies]

H2SO4 + 2KOH ---> K2SO4 + 2H2O

n(H2SO4) = 0.120 x 0.2 = 0.024

n(KOH) = 0.065 x 0.08 = 0.0052

Now, with 0.0052 moles KOH, there should be 0.0052/2 = 0.0026 moles H2SO4 (note mole ratio). However, we have 0.024 moles H2SO4. Therfore, the amount of H2SO4 in excess is 0.024-0.0026 = 0.0214 moles. Therefore, the pH of the solution is -log[0.0214 x 2] = 1.369, but pretty sure I have to make it 1, due to 0.2 being given as one significant figure...

NOTE, I multiplied 0.0214 by 2 as H2SO4 is a diprotic acid.

 

[kyokaku92]

I know it's a stupid Q but why do you need to multiply by 2 for diprotic acids?

 

[AJ92]

coz they give 2 protons for every compound (ie they increase the H+ concentration by twice as much)
similarly, you need to multiply by 3 for triprotic such as h3po4

 

[kyokaku92]

cheers for the help!!