Question

What is the pH of a mixture of 15 mL of 0.25 M HC2H3O2 and 0.46 g of NaC2H3O2 * 3H2O. Ka= HC2H3O2= 1.76*10^-5 (Please show your work)

Answer 1

This solution is a weak acid (acetic acid) with its conjugate base (sodium acetate). It is a buffer.

n of acetic acid= 15 x 0.25/1000= 0.00375 mol

since total volume is 15 and separate volume of sodium acetate is not mentioned il take 15 ml for sodium acetate as well.

n= 0.46g/ 136.03 g ( NaC2H3O2 * 3H2O) = 0.00375

n of sodium acetear= 15 x 0.0033/1000= 0.000049 mol

ph= pka + log n acetic acid/ n sodium acetate

- log Ka = pKa
pka= 4.75

ph= 4,75 + log 0.00375/ 0.000049

= 4.75+ 1.8

= 6.63

this is what im upto. please check with your teacher. please do comment if its wrong. il recheck and post it again.