Question

What is the pH of the resulting solution when 19.84 mL 0.128 M Ba(OH)2 is added to 52.05 mL 0.1267 M HCl? Enter your answer to three significant figures.

Answer 1

The answer is given in the attachment.
Answer is added 19.84 ml 52.05 m 0.128 M Ba(OH)2 0.1267 MHO. Ba(OH) 2 ² Ba 2+ + 204- milimole of Olt - = 2x 19.84 400 128=5.08 milimole of t t = 52.05 0.1267 = 6.59 excess of H+ = 6.59-5.08 = 1.56 milimolo Total rolms 71.89 m2 - Conc of H+ = [H] = 0.02 M -lug [144] = -log(@º02) or pH.70 1,51 71.89 0.02m

so pH of resulting solution 1.70.