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Using a 0.20 M phosphate buffer with a pH of 7.4, you add 0.71 mL of...

Using a 0.20 M phosphate buffer with a pH of 7.4, you add 0.71 mL of 0.45 M HCl to 51 mL of the buffer. What is the new pH of the solution?

Expert Solution

Since this is a buffer

near pH = 7.4, we need to find pKa similar to that

th enoly pKa similar to that is 7.21 which correspond to the second dissociation

H2PO4- <--> H+ and HPO4-2 ... Ka2

This is a buffer so you may used henderson hasselbalch equations:

pH = pKA + log(HA-2 / H2A-)

substitute all data you know

7.4 = 7.21 + log(HA-2 / 0.20 )

HA -2 = 0.3097 M of conjugate

NOTE that thisis before adding any HCL

When adding HCl, this will happen:

H+ ions will increase, the new euquilibrium will be:

7.4 = 7.21 + log(HA-2 / 0.20 )

HA-2 will remain the same but 0.2 will increase (now there are more H+ ions)

HA-2 = 0.3097

But H2A = 0.20 + mols of new acid

mols of new acid = M*V = 0.45*51/1000 = 0.02295

H2A new = 0.2 + 0.02295 = 0.22295

Substitute in pH equation

pH = pka + log(HA-2 / H2A-)

pH =7.21 + log(0.3097 /0.22295)

pH = 7.35

which makes sense since there is addition of acid, therefore expect lowering the pH

this is abuffer and therefore there is no huge changes on pH

queen_honey_blossom answered 9 months ago

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Using a 0.20 M phosphate buffer with a pH of 7.4, you add 0.71 mL of...

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