How would the pH of this solution be calculated?

Question

Calculate the pH of the solution that results from the addition of 0.020 moles of HNO3 to a buffer made by combining
0.500 L of 0.290 M HC3H5O2 ( use Ka = 1.30 x 10 -5 )
and 0.500 L of 0.580 M

thanks!!!!

Answer 1

Boy, "0.500 L of 0.580 M" of what??
Now, assume this "0.500 L of 0.580 M" is of CH3CH2COONa. Thus before 0.020 moles of HNO3 is added:
CH3CH2COOH: 0.145 mole
CH3CH2COONa: 0.290 mole
After 0.020 moles of HNO3 is added, the reaction:
HNO3 + CH3CH2COONa ==> NaNO3 + CH3CH2COOH is completed. Thus we have:
CH3CH2COOH: 0.165 mole
CH3CH2COONa: 0.270 mole
Notice the volume after chemical mixing is 1.000 L. So we have chemical concentrations:
CH3CH2COOH: 0.165 M
CH3CH2COONa: 0.270 M
NaNO3: 0.020 M (spectators we do not need to worry about)
Based on given:
CH3CH2COOH <==> H+ + CH3CH2COO-, Ka = 1.30x10^-5
Ka = 1.30x10^-5 = [H+]*[CH3CH2COO-]/[CH3CH2COOH]
= [H+]*(0.270/0.165)
pH = -log([H+]) = -log(Ka*0.165/0.270)
= 5.10