In 100 L solution H
2SO
4 present = 9.8 gm
In 10 L solution H
2SO
4 present =
9.8×10010 gm
In 10 L solution moles of H
2SO
4 present =
989.8×10010
In one molecule of H
2SO
4 two H
+ ion present.
In 10 L solution moles of H
+ present = 2
989.8×10010 = 0.02 moles
Also In 100 L solution NaOH present = 4 gm
In 40 L solution NaOH present =
4×10040
In 40 L solution moles of NaOH present =
404×10040
In one molecule of NaOH one OH
- ion present.
In 40 L solution moles of OH
- ion present =
404×10040 = 0.04 moles
As moles of OH
- ion is more than H
+ ion, so solution is basic.
Final Conc. of OH
– =
40+100.04−0.02 = 4
10
-4
pOH = – log (4 ×10
–4) = 3.4
pH = 14 – 3.4 = 10.6
