0.023 ~g of sodium metal is reacted with 100 ~cm ³ of water. The pH of the resulting solution is

Step-by-step solution

Given, 0.023 23 mol 100 22400 ~mol =1 10⁻³ ~mol =4.46 10⁻³ ~mol Thus, Na is the limiting reagent and decide the amount of NaOH formed. 1 mole Na give NaOH =1 ~mol 1 10⁻³ mole Na will give NaOH =1 10⁻³ ~mol Concentration of aligned [ OH ⁻ ] &= 1 10⁻³ 1000 100 =1 10⁻² pOH &=- [ OH ⁻ ] &=- (1 10⁻² ) &=2 pH &=14-2=12 aligned