Submit ous Ans uest A biochemical reaction takes place in a 1.00 mL solution of

Question

Submit ous Ans uest A biochemical reaction takes place in a 1.00 mL solution of 0.0250 M phosphate buffer initially at pH = 7.43. Incorrect, Try Again: 4 attempts remaining Acid (Proton Donor) Conjugate Base (Proton Acceptor) Ha PO4 PhosphoricDihydrogen +H+ 2.14 7.24 x 103 acid H2PO4 phosphate ion HPO4 Part B H2PO4 Dihydrogen Monohydrogen H6.86 1.38 x 10-7 phosphate ion phosphate ion During the reaction, 3.40 mol of HCl are produced. Calculate the final pH of the reaction solution. Assume that the HCl is completely neutralized by the buffer. Express your answer using two decimal places. HPO4- Monohydrogen phosphate ion PO4 Phosphate ion | 12.4 | 3.98x 10-13 Submit Re Ans

Explanation / Answer

Given
Number of mole of HCl = 3.4 x 10^-6 mol

Volume of the solution = 1 x 10^-3 L

concentration of H2PO4^- = 0.025 M

Initial moles of H2PO4^- = 0.025 x 10^-3
= 25 x 10^-6 mol

Buffer reacts with pH value pH +- 1 range

so, the reaction is

HPO4^2- + HCl -----> H2PO4^-

intially we have 25 x 10^-6 mol (HPO4^2-) present in the solution and

3.4 x 10^-6 mol of HCl present in the solution to form H2PO4^-

after buffer HCl completely nutralises that means only 3.4 x 10^-6 mol of H2PO4^- produced

so the final number of mole of HPO4^2- = (25-3.4)x 10^-6 mol = 21.8 x 10^-6 mol

final number of mole of H2PO4^- = (25+3.4) x 10^-6 mol= 28.4 x 10^-6 mol

pH = pKa + log [base]/ [acid]

= pKa + log [HPO4^2-]/ [H2PO4^-]
= pKa + log [21.8 x 10^-6 mol / volume] / [ 28.4x 10^-6 mol / volume]
= 6.86 + log [21.8/28.4]
= 6.86 -0.2
= 6.66

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