Calculate the pH of a solution prepared by mixing 2.0mL of a strong acid (HCI) solution of pH 3.0 and 3.0mL of a strong base (NaOH) of pH 10.0. (Take:log1034=1.5)
A
2.5
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B
3.5
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C
4.5
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D
6.5
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Solution
The correct option is B3.5 Number of milliequivalents of HCl=2×10−3
We know that pH+pOH=14⇒10+pOH=14pOH=10[OH−]=10−4 M3 mL of 10−4 MNaOH=3×10−4
Number of milliequivalents of NaOH=3×10−4
The resulting solution is acidic. [H+]=(2×10−3−3×10−4m. eq.)(2+3)mL=3.4×10−4MorpH=−log[H+] pH=−log(3.4×10−4)=3.5