Adding a Strong Acid to a Buffer Strong Acid to a Buffer and how buffers use res

Question

Adding a Strong Acid to a Buffer Strong Acid to a Buffer and how buffers use reserves of conjugate onjugate base to counteract the effects of se addition on pH. mixture of a conjugate acid base pair. In ds, it is a solution that contains a weak acid njugate base, or a weak base and its acid. For example, an acetic acid buffer facetic acid, CH3COOH, and its base, the acetate ion CH3COO ions cannot simply be added to a gate base is added ina salt form (eg betate NaCH3COO). ork because the conjugate acid-base pair other to neutralize the addition of H+ or s. Thus, for example, if H ions are added etate buffer described above, they willbe moved from solution by the reaction of H Ht CH3Coo CH3COOH any added OH ions will be neutralized by n with the conjugate acid: CH3COOH- CH3COO H20 Part A A beaker with 125 mL of an acetic acid buffer with a pH of 5000 is sitting on a benchtop. The total molanty of acid and conjugate base in this butter is 0.100 M. A student adds 5.70mLofa0410 MHCI solution to the beaker. How much wilthe pH change? The pKa of acetic acid is 4.740 Express your answer numerically to two decimal places. Use a minus sign if the pH has decreased. ApH Hints My Answers Give Up Review Part m a

Explanation / Answer

A beaker with 125 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 5.7 0mL of a 0.410 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740

Calculate the molarity of acid and conjugate base in the original buffer:
Use the Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
5.00 = 4.74 + log ([salt]/[acid])
log ([salt]/[acid]) = 5.00-4.74 = 0.26
[salt / [acid] = 10^0.26
[salt]/[acid] = 1.819

The total molarity of the buffer = 0.1M
Molarity of conjugate base = 0.0635M
Molarity of acid = 0.0365 M

Moles of conjugate base in 125mL of buffer = 125/1000*0.0635 = 0.00794 mol
Moles of acid in 125mL of buffer = 125/1000*0.0365 = 0.00456 mol
Mol HCl in 5.70mL of 0.390M HCl solution = 5.70/1000*0.390 = 0.00230 mol HCl

This will react with the conjugate base to produce 0.0022230 mol of CH3COOH .

The final moles of conjugate base= 0.00794-0.00222 = 0.00572 mol CH3COO-

The final moles of CH3COOH =0.00456+0.00230 = 0.00686 mol CH3COOH

pH = pKa + log([salt]/[acid])
pH = pKa + log (0.00572/0.00686)
pH = 4.74 + log 0.833
pH =4.74 + (-0.079)
pH = 4.66-answer

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