larutan asam asetat 0,1 M sebanyak 100ml mempunyai ph=3.jika 0,4 gram NaOH di camurkan kedalam larutan ini,tentukan ph larutan yang dibentuk..?(Mr NaOH=40)

100 mL CH3COOH 0,1 M pH = 3 + 0,4 gram NaOH (Mr = 40)

pH ... ?

*n CH3COOH = M × V
n CH3COOH = 0,1 × 100 mL/1000
n CH3COOH = 0,1 × 0,1 L
n CH3COOH = 0,01 mol

*n NaOH = gram/Mr
n NaOH = 0,4/40
n NaOH = 0,01 mol

*CH3COOH 0,1 M pH = 3
pH = 3
[H^+] = 10^-{pH}
[H^+] = 10^-3
[H^+] = √(Ka × Ma)
10^-3 = √(Ka × 0,1)
(10^-3)^2 = (√(Ka × 0,1) )^2
10^-6 = Ka × 0,1
Ka = 10^-6/0,1
Ka = 10^-5

#reaksi
... NaOH + CH3COOH => CH3COONa + H2O
M : 0,01 .......... 0,01
R : 0,01 ........... 0,01
________________---
S : . --- .............. --- .................. 0,01 ........... 0,01

*M = n / V total
M = 0,01 mol / (100 mL/1000)
M = 0,01 mol / 0,1 L
M = 0,1

*[OH^-] = √( Kw/Ka × M )
[OH^-] = √( 10^-14/10^-5 × 0,1 )
[OH^-] = √( 10^-9 × 10^-1 )
[OH^-] = √( 10^-10 )
[OH^-] = 10^-5

pOH = - log [OH^-]
pOH = - log 10^-5
pOH = 5

pH = 14 - pOH
pH = 14 - 5
pH = 9