Question

What will be the pH of a solution formed by mixing 40 $c{m}^3$ of 0.1 M HCl with 10 $c{m}^3$ of 0.45 M NaOH?

• A. 10
• B. 8
• C. 5
• D. 12

Answer 1

Answer: (D)

Number of moles of HCl $=\frac{MV}{1000}=\frac{0.1\times 40}{1000}=0.004$
Number of moles of NaOH $=\frac{MV}{1000}=\frac{0.45\times 10}{1000}=0.0045$
Remaining moles of NaOH after neutralization = 0.0045 – 0.004 = 0.0005
Molarity of $O{H}^{-}$ $=\frac{0.0005}{50}\times 1000=0.01\text{M}$
So pOH $=2$
So pH = 14 – 2 = 12