Question

Must be stapled here! Objective(s): Lab Report Changes in procedure (use point form): Data 1. Molarity of HCl solution: 00492 M used Volume of HCI 6.14 me II. Temperature of Ca(OH)2 solution: — III. Volume of Ca(OH)2 Volume Trial I Trial II Trial III Final burette reading (mL) 10.00 mL 10.00 mL I 10.00ml Initial burette reading (mL) Volume of Ca(OH)2 used (mL) 0:00 ml 0.00 me I 0.00 me 10.00 10.00 me I 10.oome I lovome 10.00 me Average volume of Ca(OH)2 Ksp of Ca(OH)2

Ksp of calcium hydroxide

Answer 1

The chemical equation:

Ca(OH)₂ ⇌ Ca²⁺ + 2OH¯ eqn 1

and The corresponding K_sp expression is :

K_sp = [Ca²⁺] [OH¯]²

volume of Hcl required to neutalize is (V_HCl)=6.14ml

molarity of Hcl used (M_HCl)=0.0492 M

volume of Ca(OH)₂= 10ml

every one H⁺ neutralizes one OH¯.

using M₁V₁=M₂V₂

M_HClV_HCL=MOHVOH
M_HClV_HCL/ V_OH=M_OH

6.14ml*0.0492 M/10 ml ₌M_OH

0.03M=M_OH

[OH^-1]=0.030M

[Ca²⁺]  is exactly half the [OH¯] because of the 1:2 molar ratio from the balanced equation.

[Ca²⁺] =[OH^-1]/2=0.030M/2 =0.015M

so,

K_sp = [Ca²⁺] [OH¯]²

K_sp =0.015*0.03²

K_sp =0.015*9*10^-4

K_sp =1.35*10^-5