7) berapakah ph larutan 0,1 M NaHCO3 , jika diketahui K H2CO3 = 4. 10 pangkat -7
AmaldoftYang diket hanya garam (NaHCO3) sebesar 0,1 Molar, maka terjadi sistem hidrolisis garam: [OH-] = √Kw x [G] / Ka = √10^-14 x 10^-1 / 4 x 10^-7 = √25 x 10^-10 = 5 x 10^-5
[OH-] = √Kw x [G] / Ka
= √10^-14 x 10^-1 / 4 x 10^-7
= √25 x 10^-10
= 5 x 10^-5
pOH = 5 - log 5
pH = 9 + log 5