Hey, so i got a practice sheet of questions with no answers...so i just want to know if im doing these right, thanks
- 459ml of .242 M HCL is mixed with 273 ml of .198 M Mg(OH)2. What is the pH of the resulting mixture
acid
(.459 L )(.242 M) = .111078 mol HCl
(.111078)(1 H/1HCl) = .111078 mol H
Base
(.273 L)(.198M) = .054054 mol Mg(OH)2
(.054054)(2OH/1Mg(OH)2) = .108108 OH
.111078 - .108108 = .00297 excess H
.00297/.732L = .00405 H
pH = -log(4.05) + - log(10^-3)
pH = 2.39 ?
if someone could let me know if i did the steps right that'd be awesome, thanks!