Answer to Question #143528 in Organic Chemistry for Nathaniel Whetstine

Question

What volume in milliliters of 0.0170 M Ca(OH)₂ is required to neutralize 75.0 mL of 0.0300 M HCl?

Answer 1

Required volume of Ca(OH)2=75*0.0300/(2*0.0170)=66.17ml,

Using ( V1S1=V2S2) formula,V1=volume of HCL=75mL,S1=Concentration of HCL=0.0300(M)

V2=volume of Ca(OH)2,

S2= concentration of Ca(OH)2=2*9.0170(M)