A. The pKa value for H 2 PO 4 - is 7.21 . What mole ratio of Na 2 HPO 4 to NaH 2

Question

A. The pKa value forH2PO4- is7.21. What mole ratio ofNa2HPO4 toNaH2PO4 is needed to preparea buffer with a pH of 7.41?

[HPO42-] = [H2PO4-]
B. The pKa value forH2C2O4 is1.23. What mole ratio ofKHC2O4 toH2C2O4 is neededto prepare a buffer with a pH of 1.55?

[HC2O4-] = [H2C2O4]
C. A buffer solution made from HNO2 andKNO2 has a pH of 3.72.If pKa for HNO2 is3.35, what is the[NO2-] /[HNO2] in the buffer?

[NO2-] = [HNO2]
[HPO42-] = [H2PO4-] [HPO42-] [H2PO4-] [HC2O4-] = [H2C2O4] [HC2O4-] [H2C2O4] [NO2-] = [HNO2] [NO2-] [HNO2] [HPO42-] = [H2PO4-]

Explanation / Answer

we know that for a buffered solution we have pH = pKa + log([salt]/[acid]) here we have given the values of pH and pKa so we plugin the valuesto get the ratio of concentration of salt and acid A. log( HPO42-/H2PO4) = pH - pKa                                         = 7.41 - 7.21                                         = 0.20 so HPO42-/H2PO4 = 100.2                                  = 1.584 B: log( HC2O4/H2C2O4) = pH - pKa                                         = 1.55 -1.23                                         = 0.32 so HC2O4/H2C2O4 = 100.32                                  = 2.089 C:log( NO2/HNO2) = pH - pKa                                = 3.72 -3.35                                = 0.37 so NO2/HNO2 = 100.37                            = 2.344

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