tentukan pH dari 0,69 gr CH3 COOH dilarutkan di air 250 ml ka= 1.10-^5

M asam = massa / (volume x Mr)
Mr CH3COOH = 60

[tex][H+] = \sqrt {K_a \times M_{asam}}[/tex]
[tex][H^+] = \sqrt {10^{-5} \times \frac {massa}{volume \times Mr}}[/tex]
[tex][H^+] = \sqrt {10^{-5} \times \frac {0,69}{0,25 \times 60}}[/tex]
[tex][H+] = \sqrt {4,6 \times 10^{-7}}[/tex]
[H+] = 6,78 x 10⁻⁴ M
pH = 4 - log6,78