A mixture is prepared by adding 50.0 mL of 0.200 M NaOH to 75.0 mL of 0.100 M NaOH. What is the [OH–] in the mixture?
Amount of 0.2M NaoH acid `= (0.2/1000)xx50 =0.01mol`
Amount of 0.1M NaoH acid `= (0.1/1000)xx75 = 0.0075mol`
Total NaoH moles in the solution `= 0.01+0.0075 = 0.0175mol`
Volume of the Solution `=50+75 =125 ml`
`NaOH rarr Na^++OH^-`
So now we have 0.0175moles of `OH^-` in 125ml of a solution.
`[OH^-] = (0.0175/125)xx1000 =0.14M`
So the concentration of `OH^-` is 0.14M