A buffer is prepared by adding 150 mL of 1.0 M NaOH to 250 mL of 1.0 M NaH₂PO₄. How many moles of HCl must be added to this buffer solution to change the pH by 0.18 units? If necessary, assume the total volume remains unchanged at 400 mL. A)0.025 mol HCl B)0.063 mol HCl C)0.082 mol HCl D)0.50 mol HCl E)1.0 mol HCl

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The correct answer is A because the change in pH is small, so the Henderson-Hasselbalch equation can be used to approximate the number of moles of HCl that must be added. The Henderson-Hasselbalch equation is:```pH = pKa + log([A-]/[HA])```where pKa is the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.In this case, the pKa of NaH2PO4 is 7.21. The initial concentration of NaH2PO4 is 0.25 M, and the initial concentration of NaOH is 0.15 M. Therefore, the initial concentration of H2PO4- is:```[H2PO4-] = [NaH2PO4] - [NaOH] = 0.25 M - 0.15 M = 0.10 M```The initial concentration of HPO42- is:```[HPO42-] = [NaOH] = 0.15 M```The change in pH is 0.18 units, so the new pH will be:```pH = 7.21 + 0.18 = 7.39```Therefore, the new concentration of H2PO4- will be:```[H2PO4-] = 10^(-(pH - pKa)) = 10^(-(7.39 - 7.21)) = 0.050 M```The new concentration of HPO42- will be:```[HPO42-] = [NaH2PO4] - [H2PO4-] = 0.25 M - 0.050 M = 0.20 M```The number of moles of HCl that must be added to change the pH by 0.18 units is:```moles of HCl = [H2PO4-] - [HPO42-] = 0.050 M - 0.20 M = -0.15 M```The negative sign indicates that HCl must be added to the buffer solution. Therefore, the number of moles of HCl that must be added is 0.15 mol.

A Buffer Is Prepared by Adding 150 ML of 1