What is the pH of a 0.100 M solution of the salt sodium formate (NaCHO2)? The Ka

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[HCHO2] = 0.12 / 1.90 =0.0632 M [CHO2-] = 0.10 / 1.90 = 0.0526 M pKa = - log Ka = 3.74 pH = pKa + log [CHO2-] / [HCHO2] pH = 3.74 + log 0.0526 / 0.0632 = 3.66 this is the first HCHO2 + OH->> CHO2- + H2O Moles HCHO2 = 0.12 - 0.02 = 0.10 Concentration HCHO2 = 0.10 / 1.90 =0.0526 M Moles CHO2- = 0.10 + 0.02 =0.12 Concentration CHO2- = 0.12 / 1.90 = 0.0632 CHO2- + H+ >> HCHO2 Moles CHO2- = 0.10 - 0.04 =0.06 concentration CHO2- = 0.06 / 1.90 =0.0316 Moles HCHO2 = 0.12 + 0.04 =0.16 concentration HCHO2 = 0.16 / 1.90 =0.0842 M pH = 3.74 + log 0.0316 / 0.0842 =3.31 I hope this help you pH = 3.74 + log 0.0632 / 0.0526 =3.82

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