Calculate the number of moles of HCl( g ) that must be addedto 1.0 L of 1.0 M Na

Question

Calculate the number of moles of HCl(g) that must be addedto 1.0 L of 1.0 MNaC2H3O2 to produce a solutionbuffered at each pH. (a) pH = pKa
so i thought i knew how to do this, but i got the answer wrong. Theonly place I can understand how I got it wrong is with which Kavalue I should use (maybe I dont have the right Ka value?) Thanks :) (a) pH = pKa
so i thought i knew how to do this, but i got the answer wrong. Theonly place I can understand how I got it wrong is with which Kavalue I should use (maybe I dont have the right Ka value?) Thanks :)

Explanation / Answer

We Know that According to Henderson'sequation :        pH = pKa + log [ salt ] / [ Acid ]         Number of moles= Molarity x volume of solution in L                                   =   1 M x 1 L                                                           =   1 mole           CH3COONa +   HCl  -------->   CH3COOH + NaCl           From the above balanced equation it is clear that 1 moleof salt can be hydrolysed in the presence of 1 moleof             Acid.                         Therefore we have to add 1 mole of HCl to given salt so that the pH of solution will be equal topKa.                    pH = 4.74472 + log [ 1 mole ] / [ 1 mole ]                    pH = 4.74472

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