At certain temperature, pH of 0.01N BaOH2 is 11.699. Ionic product of water a that at temperature is

OH−=10−2M;pOH=2;pH=11.699;pH+pOH=13.699;      pkW=13.699;     pkW=−logkWlog kw=−13−0.699log kw=−13−0.699+1−1log kw=−14+0.301kw=Antilog14¯.3010kw=2×10−14