Berapakah ph larutan jika 10 g NaOH dilarutkan ke dalam 250 ml HCL 0,5M?

Diketahui :
m NaOH = 10 gram
Volume larutan HCl = 250 ml = 0,25 L
Molaritas HCl = 0,5 M = 0,5 mol/L

Ditanya : pH larutan setelah NaOH dilarutkan ke dalam larutan HCl ?

Jawab :

*Mencari mol NaOH :

Mr NaOH :
= Ar Na + Ar O + Ar H
= 23 + 16 + 1
= 40 gram/mol

mol NaOH :
= m / Mr
= 10 gram / 40 gram/mol
= 0,25 mol

*Mencari mol HCl :

mol HCl :
= M . V
= 0,5 mol/L . 0,25 L
= 0,125 mol

*Mencari mol tersisa :
______NaOH + HCl —> NaCl + H2O
awal :0,25____0,125______________
reaksi:0,125__0,125______________
akhir :0,125____-_________________

mol tersisa adalah NaOH sebanyak 0,125 mol , sehingga larutan bersifat basa.

[OH-] :
= n / V
= 0,125 mol / 0,25 L
= 0,5 M
= 5.10-¹ M

pOH :
= – log [OH-]
= – log 5.10-¹
= – ( log 5 + log 10-¹)
= – log 5 – log 10-¹
= 1 – log 5

pH :
= 14 – pOH
= 14 – ( 1 – log 5)
= 14 – 1 + log 5
= 13 + log 5