vetchem Posted April 7, 2008 Share Posted April 7, 2008 I think it's the wording of the question that has me so hung up on this question... A 200 mL NaOH solution was added to 400 mL of a 2.00 M HNO2 (Ka = 4.5 x 10-4). The pH of the resulting solution was 1.5 units greater than that of the original acid solution. Calculate the molarity of the NaOH solution. Can someone help? Link to comment Share on other sites More sharing options...
hermanntrude Posted April 7, 2008 Share Posted April 7, 2008 OK so calculate the pH of the original solution. Then you'll know what the pH of the mixture was. Then you can work backwards to find the number of moles of hydroxide required to change the pH, then you can finally calculate the concentration of the original mixture. An ICE table will help, too. Link to comment Share on other sites More sharing options...
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