The ph of a solution that contains 0.818 m acetic acid (ka = 1.76 x 10-5) and 0.172 m sodium acetate is

PH is calculated using Handerson- Hasselbalch equation,

                                      pH  =  pKa  +  log [conjugate base] / [acid]

Conjugate Base  =  Acetate (CH₃COO⁻)
Acid                    =  Acetic acid (CH₃COOH)
So,
                                      pH  =  pKa  +  log [acetate] / [acetic acid]

We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,

                                     pKa  =  -log Ka
Putting value of Ka,
                                     pKa  =  -log 1.76 × 10⁻⁵

                                     pKa  =  4.75
Now,
Putting all values in eq. 1,
                     
                                     pH  =  4.75 + log [0.172] / [0.818]

                                     pH  =  4.072

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