Hitung ph dari 200 ml larutan yang mengandung 0.049 mg asam sulfat mr : 98

Diketahui :
Volume larutan = 200 ml = 0,2 L
Mr H2SO4 = 98 gram/mol
m H2SO4 = 0,049 mg = 0,000049 gram

Ditanya : pH larutan ?

Jawab :

mol H2SO4 :
= m / Mr
= 0,000049 gram / 98 gram/mol
= 5.10^-7 mol

Molaritas H2SO4 :
= n / V
= 5.10^-7 mol / 0,2
= 2,5 × 10^-6 M

[H+] :
= a . M
= 2 . 2,5 × 10^-6 M
= 5 × 10^-6 M

pH :
= – log [H+]
= – log 5 × 10^-6
= – ( log 5 + log 10^-6)
= – log 5 – log 10^-6
= 6 – log 5
= 6 – 0,69
= 5,31

Coba deh di gambar. Cari dulu molnya trus M nya