500 ml asam lemah HA memiliki pH 6. jika Ka HA = 1x10 ^ -10, massa HA (Mr = 27) yang terlarut ..... g

pH=6
[H+]=10^-6
[H+]=√(Ka.M)
10^-6=√(10^-10 . M)
10^-12=10^-10 . M
M=10^-2

M=(w/mr)×(1000/l)
10^-2=(w/27)×(1000/500)
10^-2=2w/27
27×10^-2=2w
w=13,5 × 10^-2 g
=0,135 g