JAMB - Chemistry (2015 - No. 54)
what is the pH of 0.00lmoldm3 solution of sodium hydroxide
14
13
12
11
Explanation
[H+][OH−] = 10 − 4
[H+] =10 − 14 ÷ [OH+]
=10 −14 ÷ (1 × 10 − 3) = 10 − 11
pH = − log10[H+ ]
= − log10[10 − 11] = − 1 × −11
log10[10 − 11] = 1
= − 1 × 11 × 1
= − 11
[H+] =10 − 14 ÷ [OH+]
=10 −14 ÷ (1 × 10 − 3) = 10 − 11
pH = − log10[H+ ]
= − log10[10 − 11] = − 1 × −11
log10[10 − 11] = 1
= − 1 × 11 × 1
= − 11
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