MCQ
What is $\ce{pH}$ of resulting solution when equal volume when equal of $0.1M ,\ce{NaOH}$ and $0.01M, \ce{HCl}$ are mixed? $[\log 4.5 = 0.65]$
- A$7$
- B$1.04$
- ✓$12.65$
- D$2.0$
✓
Answer
Correct option: C.
$12.65$
Number of mole of $[\text{OH}^-]=\frac{0.1}{2}=0.05$ on mixing
$[\text{H}^+]=\frac{0.01}{2}=0.005$
No. of mole of $\ce{[OH}^-] $ left $= 0.05 - 0.005 = 0.045$
$\text{pOH} = -\log 4.5 \times 10-2$
$= -0.65 + 2.00 = 1.35$
$\ce{pH } = 14 - 135 = 12.65$
$[\text{H}^+]=\frac{0.01}{2}=0.005$
No. of mole of $\ce{[OH}^-] $ left $= 0.05 - 0.005 = 0.045$
$\text{pOH} = -\log 4.5 \times 10-2$
$= -0.65 + 2.00 = 1.35$
$\ce{pH } = 14 - 135 = 12.65$
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