Question

Calculate the pH and pOH of the following solutions:

a) 1:1 volume ratio mixture of 0.0125 M HCl and 0.0125 M Ca(OH)₂

b) 2:3 volume ratio mixture of 0.0125 M HNO₃ and 0.0125 M KOH

Answer 1

a) 1:1 volume ratio mixture of 0.0125 M HCl and 0.0125 M Ca(OH)2

is 1:1 ratio then, 1+1 = 2, so this is 50%-50%

molaritie sare halved

0.0125/2 = 0.00625 mol of HCl

0.0125/2 = 0.00625 mol of Ca(OH)2

ratio is 1:2 since

Ca(OH)2 + 2HCl = 2H2O + CaCl2

2*0.00625 -0.00625 = 0.00625 M

[OH-] = 0.00625 M

pOH = -log([OH-]) = -log(0.00625)= 2.204

pH = 14-pOH = 14-2.204

pH = 11.796

b)

if 2:3 ratio

V total = 2+3 = 5

2/5 * 0.0125 = 0.005

3/5*0.0125 = 0.0075

KOH + HNO3 = H2O + KNO3

raito is 1:!

0.0075 - 0.0050 = 0.0025

then

[OH] = 0.0025

pOH = -log([OH-]) = -log(0.0025) = 2.602

ph = 14-pOH

pH = 11.398