Question

Terdapat 200ml larutan H2SO4 yang memiliki pH=1. berapa jumlah NaOH (Mr=40) padatan yang harus ditambahkan agar pH larutan berubah menjadi 2? perubahan volume akibat penambahan NaOH dianggap tidak terjadi

Answer 1

PH = - log [H^+]
1 = - log [H^+]
[ H^+ ] = - antilog 1
[ H^+ ] = 10^-1

H2SO4 -> 2H^+ + SO4^-2

[H^+] = n x M
10^-1 = 2 x M
M = 5 x 10^-2

pH' = - log [H^+]'
2 = - log [H^+]
[H^+] = 10^-2

H2SO4 + NaOH -> Na2SO4
+ H20

M' = 10^-2

Mc = M x V + M' x V / 2V
= 5 x 10^-2 + 10^-2 /2
= 6 x 10^-2 / 2
= 3 x 10^-2

Mc = g / Mr x 1000 / V
3 x 10^-2= g/40 x 1000/200
3 x 10^-2 = 5g / 40
5g = 1,2
g = 0,024 gram