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Question

Chemistry Question on Equilibrium

A solution of an acid has pH = 4.70. Find out the number of OH−OH^- ions (pKw=14)(pK_w = 14)

A

5×10−105 \times 10^{-10}

B

4×10−104 \times 10^{-10}

C

2×10−52 \times 10^{-5}

D

9×10−49 \times 10^{-4}

Answer

5×10−105 \times 10^{-10}

Explanation

Solution

pH=4.70⇒−log[H+]=4.70pH=4.70 \Rightarrow -log\left[H^{+}\right]=4.70 ∴log[H+]=−4.7=5ˉ⋅30\therefore log \left[H^{+}\right]=-4.7=\bar{5}\cdot30 or [H+]=1.999×10−5=2×10−5\left[H^{+}\right]=1.999 \times10^{-5}=2\times10^{-5} pKw=14⇒[H+]×[OH−]=10−14pK_{w}=14 \Rightarrow \left[H^{+}\right]\times\left[OH^{-}\right]=10^{-14} ∴[OH−]=10−142×10−5=0.5×10−9\therefore\left[OH^{-}\right]=\frac{10^{-14}}{2\times10^{-5}}=0.5\times10^{-9} =5×10−10=5\times10^{-10}