Question

What is the ph of a buffer prepared by adding 0.506 mol of the weak acid ha to 0.609 mol of naa in 2.00 l of solution? the dissociation constant ka of ha is 5.66×10−7. express the ph numerically to two decimal places?

Answer 1

Hello!

To solve this question, we first have to know the pKa of the acid:


`pKa=-log(Ka)=-log(5,66*10^(-7))=6,25`

Now, we can apply the Henderson-Hasselbach's equation to determine the pH of the buffer solution:


`pH=pKa+ log( ([NaA])/([HA]) )= 6,25 + log(((0,609 mol/2L))/((0,506 mol/2L)))=6,33`

So, the pH of this buffer solution is 6,33

Have a nice day!