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Q.
What will be the pH of a solution formed by mixing 40 $cm^{3}$ of 0.1 M HCl with 10 $cm^{3}$ of 0.45 M NaOH?
NTA AbhyasNTA Abhyas 2022
Solution:
Number of moles of HCl $=\frac{M V}{1000}=\frac{0.1 \times 40}{1000}=0.004$
Number of moles of NaOH $=\frac{M V}{1000}=\frac{0.45 \times 10}{1000}=0.0045$
Remaining moles of NaOH after neutralization = 0.0045 – 0.004 = 0.0005
Molarity of $OH^{-}$ $=\frac{0.0005}{50}\times 1000=0.01\text{M}$
So pOH $=2$
So pH = 14 – 2 = 12