What is the pH of a solution of 0.15 M formic acid?

Having trouble grasping these problems. Calculate the pH of a Solution with 32.9g Formic acid HCOOH per Liter?

• now I calculated the Molarity to be 0.174 M I THINK thats the initial because it comes from the 32.9g and I dont beleive this is a strong acid. So my equation is HCOOH + H2O <> H3O + HCOO- SO the initial concentrations would be HCOOH =0.174 H3O= 0 and HCOO- = 0 So How do I solve? I dont have a Ka value so not sure i can use the quadratic way.

• Answer:

  Since it's a weak organic acid-You must use the Ka value and mass action expression to find [H+] Ka=[H+][COOH-]/[HCOOH] 32.9g HCOOH * (1 mol/46g)=0.7152 mol HCOOH Since it's in 1L solution Mols=molarity Ka=1.8 x 10–4 so 1.8 x 10–4=[H+][COOH-]/0.7152 [H+] and [COOH-] are the same values, because 1 molecule of HCOOH produces 1 of each. So set it as X^2 1.8x10^-4=x^2/0.7152 1.28x10^-4=x^2 Take square root x=[H+]=[COOH-]=0.011346188 Now, pH=-log[H+] -log[0.011346188]=1.94515301