What volume, in mL, of 0.23 M HCl neutralizes 17.72 mL of 0.22 M Ca(OH)2 [Solved]

What quantity, in mL, of 0.23 M HCl neutralizes 17.72 mL of 0.22 M Ca(OH)2

Moles Ca(OH)2 = 0.01772 L x 0.22 M = 0.0039
Moles OH- = 2 x 0.0039 = 0.0078
= moles HCl
V = moles / M = 0.0078 / 0.23 =0.034 L
=> 34.0 mL

This post is last updated on hrtanswers.com at Date : 1st of September – 2022