Problem: Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid.

FREE Expert Solution

The dissociation of a monoprotic weak acid (HA) is as follows:

HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq)


From this, we can construct an ICE table. Remember that liquids are ignored in the ICE table and Ka expression.

View Complete Written Solution
Problem Details

Enough of a monoprotic acid is dissolved in water to produce a 0.0192 M solution. The pH of the resulting solution is 2.53. Calculate the Ka for the acid.

All Chemistry Practice Problems Weak Acids Practice Problems
Q.In a 1.0 x 10-4 M solution of HCN(aq), identify the relative molar amounts of these species. Arrange from most to least. H2OH3O+HCNOH-CN-
Q.Which of the following is the equilibrium constant expression for the dissociation of the weak acid HOCl?B) K = [H+][OCl–]E) none of these
Q.If an acid, HA, is 10.0% dissociated in a 1.0 M solution, what is Ka for this acid?A) 6.3 × 10–2B) 9.1 × 10–2C) 8.1 × 10–1D) 1.1 × 10–2E) none of thes...
Q.Calculate the pH of 0.100 M sulfurous acid, H 2SO3. (Ka1 = 1.72 x 10 -2, Ka2 = 6.43 x 10 -8)
See all problems in Weak Acids

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Weak Acids concept. If you need more Weak Acids practice, you can also practice Weak Acids practice problems.