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maaf saya copas pertanyaan kamu :)
HCOOH = asam lemah
HCOONa = garam
gunakan rumus
pH = pKa - log (M asam / M garam)
5 = - log ka - log (M asam / M garam)
5 = - log 10^-6 - log (0,1 / M garam)
5 = 6 - log (0,1 / M garam)
5-6 = - log (0,1 / M garam)
-1 = - log (0,1 / M garam)
1 = log (0,1 / M garam)
10 = 0,1/Mgaram
M garam = 0,01 M
karena disini tidak diketahui volume larutan maka kita anggap 0,01 M = 0,01 mol
maka massa HCOONa
= mol x Mr HCOONa
= 0,01 x 68
= 0,68 gram
semoga membantu :)
beri label terbaik ya :D
(tombolnya ada di kanan bawah :D)