to 25ml of 023 molar ammonium hydroxide solution 25 ml of 0.25 M HCl is added. What will be the ph of the resulting solution?

Question

i am struggling to figure out to to get ph using the kb value for ammonium hydroxide of 1.8x10^-5. 

Answer 1

The amount of base present at start is (25 mL)(0.23 M) = 5.75 mmoles

The amount of acid added is (25 mL)(0.25 M) = 6.25 mmoles

After the two solutions are mixed, the pH will be governed by the amount of excess H⁺, which is 0.5 mmoles/(50 mL) = 0.01 M.  The pH of the solution will be -log(.01) = 2

 

The K_b value for ammonium hydroxide is not needed to answer this problem.