pH = pKa + log [A-]/[HA] 1. What is the pH of a solution containing 0.042 M NaH2

Question

pH = pKa + log [A-]/[HA] 1. What is the pH of a solution containing 0.042 M NaH2PO4 and 0.058 M Na2HPO4 ? The pKa of sodium phosphate is 6.86. 2. What is the pH of the sodium phosphate buffer described in problem #1 if 1 ml of 10M NaOH is added? What is the pH of 1 liter of water if 1ml of NaOH is added? How does this illustrate the importance of buffers? 3. What is the pH of a solution prepared by mixing 200 ml of 0.1 M HCl with 300 ml of 0.1 M sodium acetate and diluting the mixture to 1 L? The pKa of acetic acid is 4.76 4. Calculate the ratio of dihydrogen phosphate to hydrogen phosphate in the blood at pH 7.4. The Ka = 6.3 x 10 -8 5. For a weak acid with a pKa of 6.0 calculate the ratio of conjugate base to acid at a pH of 5.0.

Explanation / Answer

H2PO4 ...=====>...HPO4...+...H+ pH = pKa – Log ([HA]/[A-]) DATA: [NaH2PO4]=0.042-M [Na2HPO4]=0.058-M pKa=6.86 pH=6.86-Log(0.042/0.058) pH=6.86-0.140 pH=6.72 similarly other can be done

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