What is the pH of this solution?

Question

A 1.7 L solution is made with water and 0.23 g of HNO₃ (MW = 63.01 g/mol). What is the pH of this solution?

I am having a hard time with this question and would like some help. Here's what I have so far, correct me if I am approaching this wrong:

HNO₃ + H₂O = HNO₃ (aq) = H⁺ + NO^-3

Since I was given 0.23 g and 63.01 g/mol, I was hoping to do something with the moles of HNO₃. I did (0.23g)(63.01g/mol) = 14.4923 mol HNO₃.

I don't know how to proceed from here to find the pH. Please help?

Answer 1

You calculated mokes incorrectly. It is 0.23 g x 1 moke/63.01.g = 0.00635 moles HNO₃

As a strong acid this ionizes completely to give 0.00635 moke H⁺.

[H⁺] = 0.00636 mol/1.7 L = 0.0021 M

pH = -log 0.0021

pH = 2.67