(b) what is the ph of a solution containing 0.03m benzoic acid (pka 3.19) and 0.02m nabenzoate?

Answer is: pH of solution is 3,02.
c(benzoic acid) = 0,03 mol/dm³.
c(sodium benzoate) = 0,02 mol/dm³.
pKa(benzoik acid) = 3,19.
pH = ?
Henderson–Hasselbalch equation for buffers: pH = pKa + log(cs/ck)
pH = 3,19 + log(0,02/0,03)
pH = 3,19 - 0,176 
pH = 3,02.
cs - concentration of acid.

So what you would do is go through and add 3 teaspoons(tsp) 6 times because of the 6 tablespoons(tbsp). which will give you an answer of 18. another way is to multiply 3x6=18.

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