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A mixture is prepared by adding 50.0 mL of 0.200 M NaOH to 75.0 mL of 0.100 M NaOH....
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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
Posted by jeew-m on August 5, 2013 at 4:44 PM (Answer #1)
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