I have been working on this problem since last night. Istill can\'t figure it ou

Question

I have been working on this problem since last night. Istill can't figure it out. Any information will behelpful. The problem reads: What volume of 0.25M NaH2PO4 must be added to Na2HPO4- inorder to make a 100mL buffer solution with a pH of 7.40. Ka=6.3*10-8 Thank you. I have been working on this problem since last night. Istill can't figure it out. Any information will behelpful. The problem reads: What volume of 0.25M NaH2PO4 must be added to Na2HPO4- inorder to make a 100mL buffer solution with a pH of 7.40. Ka=6.3*10-8 Thank you.

Explanation / Answer

According to Henderson's equation,                     pH = pKa + log [HPO4-2] /[H2PO4-]                   7.40 = - log (6.3*10-8 ) +  log[HPO4-2] /[H2PO4-]            7.40 - 7.20 =  log [HPO4-2] /[H2PO4-]                   0.2 =  log [HPO4-2] /[H2PO4-] [HPO4-2] /[H2PO4-]   = 100.2
                                               = 1.5848       [HPO4-2] /[H2PO4-]   =   1.5848 Thus [H2PO4-] / [HPO4-2] = 1 /1.5848                                             = 0.630 /1 This indicates that the above buffer is formed when weadd the 0.630 molesof  H2PO4- forevery 1 moles of HPO4-2. If we assume that there is 1 mole ofHPO4-2 is present then, The volume of theH2PO4- required = 0.630 mole /0.25 mol/L                                                        = 2.52 mL

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