How many milliliters of 3.2 M HCl must be added to 3.1 L of 0.20 M K 2 HPO 4 to

Question

How many milliliters of 3.2 M HCl must be added to 3.1 L of 0.20 M K2HPO4 to prepare a pH = 7.45 buffer?  (See the appendix.)

Acid Name ConjugateAcid Ka pKa Conjugate Base Base Name perchloric acid HClO4 >>1 < 0 ClO41- perchlorate ion hydrohalic acid HX (X=I,Br,Cl) >>1 < 0 X1- halide ion sulfuric acid H2SO4 >>1 < 0 HSO41- hydrogen sulfate ion nitric acid HNO3 >>1 < 0 NO31- nitrate ion hydronium ion H3O1+ 1.0 0.00 H2O water iodic acid HIO3 0.17 0.77 IO31- iodate ion oxalic acid H2C2O4 5.9 x 10-2 1.23 HC2O41- hydrogen oxalate ion sulfurous acid H2SO3 1.5 x 10-2 1.82 HSO31- hydrogen sulfite ion hydrogen sulfate ion HSO41- 1.2 x 10-2 1.92 SO42- sulfate ion phosphoric acid H3PO4 7.5 x 10-3 2.12 H2PO41- dihydrogen phosphate ion hydrofluoric acid HF 7.2 x 10-4 3.14 F1- fluoride ion nitrous acid HNO2 4.0 x 10-4 3.40 NO21- nitrite ion lactic acid HC3H5O3 6.4 x 10-5 3.85 C3H5O31- lactate ion formic acid HCHO2 1.8 x 10-4 3.74 CHO21- formate ion hydrogen oxalate ion HC2O41- 6.4 x 10-5 4.19 C2O42- oxalate ion hydrazoic acid HN3 1.9 x 10-5 4.72 N31- azide ion acetic acid HC2H3O2 1.8 x 10-5 4.74 C2H3O21- acetate ion carbonic acid H2CO3 4.3 x 10-7 6.37 HCO31- hydrogen carbonate ion hydrogen sulfite ion HSO31- 1.0 x 10-7 7.00 SO32- sulfite ion hydrosulfuric acid H2S 1.0 x 10-7 7.00 HS1- hydrogen sulfide ion dihydrogen phosphate ion H2PO41- 6.2 x 10-8 7.21 HPO42- hydrogen phosphate ion hypochlorous acid HClO 3.5 x 10-8 7.46 ClO1- hypochlorite ion ammonium ion NH41+ 5.6 x 10-10 9.25 NH3 ammonia hydrocyanic acid HCN 4.0 x 10-10 9.40 CN1- cyanide ion hydrogen carbonate ion HCO31- 4.7 x 10-11 10.33 CO32- carbonate ion hydrogen phosphate ion HPO42- 4.8 x 10-13 12.32 PO43- phosphate ion hydrogen sulfide ion HS1- 1.3 x 10-13 12.89 S2- sulfide ion water H2O 1.0 x 10-14 14.00 OH1- hydroxide ion ammonia NH3 <<10-14 NH21- amide ion hydroxide ion OH1- <<10-14 O2- oxide ion

Explanation / Answer

Let y ml of HCL be added

we know that

moles = molarity x volume (ml) / 1000

so

moles of HCl added = 3.2 x y / 1000 = 3.2y x 10-3

moles of K2HP04 = 0.2 x 3.1 = 0.62

now

the reaction is

HP042- + H+ --> H2P04-

we can see that

moles of HP042- reacted = moles of HCl added = 3.2y x 10-3

moles of H2P04- formed = moles of HCl added = 3.2y x 10-3

so

finally

moles of HP042- = 0.62 - ( 3.2y x 10-3)

moles of H2P04- = 3.2y x 10-3

now

for a buffer

pH = pKa + log [conjugate base / acid ]

pH = pKa + log [ HP042- / H2P04-]

using the table

pKa for H2P04- is 7.21

so

7.45 = 7.21 + log [ 0.62 - ( 3.2y x 10-3) / 3.2y x 10-23) ]

0.24 = log [ ( 0.62/ 3.2y x 10-3) - 1]

[ ( 0.62/ 3.2y x 10-3) - 1] = 1.7378

( 0.62/ 3.2y x 10-3) = 2.7378

y = 70.77

so

70.77 ml of HCl must be added

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