4. Larutan asam sulfat 0.03 M sebanyak 100 mL dicampur dengan 200 mL larutan 0.01M HNO3.Tentukanlah pH larutan sebelum dan sesudah dicampur !
4. Larutan asam sulfat 0.03 M sebanyak 100 mL dicampur dengan 200 mL larutan 0.01M HNO3.
Tentukanlah pH larutan sebelum dan sesudah dicampur !
PEMBAHASAN
[H+] = Valensi x M
Konsentrasi [H+] H2SO = Valensi x M
[H+] H2SO4 = 2 x 0,03 M
[H+] H2SO4 = 0.06 => 6 x 10 M
pH awal H2SO4
PH = -log 6x10^ -2
PH = 2 - log 6
Konsentrasi [H+] HNO = Valensi x M
= 1 x 0.01 M
= 0.01 M
= 1x10 M
pH awal HNO3:
PH = -log [H]
PH = -log 1 x 10
PH = 2
Ditanya: pH larutan sebelum dan sesudah dicampur ?
Mol H+ dalam H2SO4:
Mol [H+] = 2 x M x V
Mol [H+] = 2 x 0.03 M x 0.1 L
Mol [H+] = 6 x 10-3
mol Mol [H+] HNO3 = 1 x M x V
[H+] = 1 x 0,01 x 0,2 L
[H+] = 2x10-3
mol Maka mol total H dalam larutan setelah dicampur adalah:
mol [H+] H2SO4 + mol [H+] HNO3
= 6x10-3 + 2x10-3
= 8x10-3 mol
Vol campuran = 100 + 200 mL
= 300 mL
[H+ campuran] = (mol campuran/volume campuran)
[ H+ campuran] = (8 x 10-3/0.3 L)
[ H+ campuran] = 0,027
[ H+ campuran] = 2.7 x10-3 M
pH campuran
PH = -log [H+campuran]
PH = -log 2,7 x 10-3
PH = 3 - log 2,7
