Chemistry 12th Edition

We are given that $[HCl]=1.4\times 10^{-3}M$ We know that $[H^+]\times [OH^-]=1\times 10^{-14}$ Thus $OH^-=\frac{1\times 10^{-14}}{H^+}=\frac{1\times 10^{-14}}{1.4\times 10^{-3}}=7.14\times 10^{-10}$