Learning Goal: The following calculations are similar to those you will complete

Question

Learning Goal:

The following calculations are similar to those you will complete as part of the lab.

Buffers work because the conjugate acid-base pair work together to neutralize the addition of H+ or OH ions. Thus, for example, if H+ ions are added to the acetate buffer described above, they will be largely removed from solution by the reaction of H+ with the conjugate base: H++CH3COOCH3COOH

Similarly, any added OH ions will be neutralized by a reaction with the conjugate acid: OH+CH3COOHCH3COO+H2O

This buffer system is described by the Henderson-Hasselbalch equation pH=pKa+log[conjugate base]/[conjugate acid]    

The pKa of acetic acid is 4.74.

Your Buffer System:

A beaker with 2.00×102 mL of an acetic acid buffer with a pH of 4.90 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M .

Part A: How many moles of acid are in this buffer solution?

Part B: How many moles of the conjugate base are in this buffer solution?

Part C: If you add 5.70 mL of a 0.470 M HCl solution to the beaker, how much will the pH change?

Explanation / Answer

pH = pKa + log(salt)/(acid)
4.9 = 4.74 + log(0.1(x)/(0.1*(200-x)))
x = 118.2 mL
Number of moles of acid = 0.1*118.2/1000 = 0.01182 moles
Number of moles of Conjugate base = 0.1 * (200-118.2)/1000 = 0.00818 moles
pH = pKa +log(salt)-(acid)/(acid)+(acid)
pH = 4.74 + log((0.00818-(5.7*0.47/1000))/(0.01182+(5.7*0.47/1000)))
pH = 4.319

Cahange in PH = 4.9 - 4.319 = 0.581

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