Calculate the pH of 0.189 M phosphoric acid (H3PO4, a triprotic acid). Ka1 = 7.5

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Explanation / Answer

H3PO4 -----------> H2PO4- + H+ (Ka1 = 7.5 x 10^-3) H2PO4- ----------> HPO42- + H+ (Ka2 = 6.2 x 10^-8) HPO42- -------------> PO43- + H+ (Ka3 = 4.8 x 10^-13) Now, For Ka1 we get conc of H+ = sqrt(Ka*c) = 3.765 x 10^-2 M We will not consider the H+ from Ka2 and Ka3 because the value of [H+] from these two will be extremely small and the sum will be equal to 3,765 x 10^-2 M So, pH = -log(3.765 x 10^-2) = 1.424

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