\(\left[OH^-\right]=0,0005.2=0,001M=10^{-3}M\)
\(\left[H^+\right].\left[OH^-\right]=10^{-14}\rightarrow\left[H^+\right]=\dfrac{10^{-14}}{\left[OH^-\right]}=\dfrac{10^{-14}}{10^{-3}}=10^{-11}\)
pH=-lg\(\left[H^+\right]\)=-lg10-11=11
- Hoặc: pH=14+lg\(\left[OH^-\right]\)=14+lg10-3=14-3=11