Propanoic acid CH3CH2COOH is a weak acid. pKa=4.89 => Ka=1.3*10-5
Use the (almost correct approximation) formula:
[H+] = SQRT [ Ka * ca] , if at least [Ka/ca]< 0.04 and pH< 6.8 (which is true in this case)
So: pH = 0.5*pKa - 0.5*log(ca) = 0.5*4.89 - 0.5*log(0.265) =2.445-(-0.288) = 2.733 = 2.7
4.50 g(1/169.88) =.0265 mols
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