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`
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