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Question
What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?
Options
2.0
3
7.0
12.65
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Solution
12.65
Explanation:
x ml of 0.1 m NaOH + x ml of 0.01 M HCI
No. of moles of NaOH = 0.1 × x × 10−3 = 0.1x × 10−3
No. of moles of HCl = 0.01 × x × 10−3 = 0.01x × 10−3
No. of moles of NaOH after mixing = 0.1x × 10−3 – 0.01x × 10−3
= 0.09x × 10−3
Concentration of NaOH = `(0.09"x" xx 10^-3)/(2"x" xx 10^-3)` = 0.045
[OH–] = 0.045
pOH = –log (4.5 × 10−2)
= 2 – log 4.5
= 2 – 0.65
= 1.35
pH = 14 – 1.35 = 12.65
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