Buffer A 0.3731g NaC2H3O2 0.1 M HC2H3O2 Buffer B 1.12g NaC2H3O2 Calculate the co
ID: 504961 • Letter: B
Question
Buffer A
0.3731g NaC2H3O2
0.1 M HC2H3O2
Buffer B
1.12g NaC2H3O2
Calculate the concentration of hydronium ion of both buffer solutions at their starting pHs. Calculate the moles of hydronium ion present in 20.0 mL of each buffer. A change of pH by 1 unit means a change in hydronium ion concentration by a factor of 10. Calculate the number of moles of NaOH that would theoretically be needed to decrease the moles of hydronium calculated in #1 by a factor of 10 for each buffer. Are their any differences between your experimental results and the theoretical calculation. Which buffer had a higher buffering capacity? Why?Explanation / Answer
Formula,
moles = grams/molar mass
molarity = moles/L of solution
1. Buffer A
molarity of NaC2H3O2 = 0.3731 g/82.03 g/mol x 0.02 L = 0.23 M
molarity of HC2H3O2 = 0. 1 M
Initial pH
pH = pKa + log(base/acid)
= 4.74 + log(0.23/0.1)
= 5.10
pH = -log[H3O+]
[H3O+] = 7.91 x 10^-6 M
In 20 ml buffer,
moles of H3O+ = 7.91 x 10^-6 M x 0.02 L
= 1.58 x 10^-7 mol
Buffer B
molarity of NaC2H3O2 = 1.12 g/82.03 g/mol x 0.02 L = 0.68 M
molarity of HC2H3O2 = 0.3 M
Initial pH
pH = pKa + log(base/acid)
= 4.74 + log(0.68/0.3)
= 5.10
pH = -log[H3O+]
[H3O+] = 7.91 x 10^-6 M
In 20 ml buffer,
moles of H3O+ = 7.91 x 10^-6 M x 0.02 L
= 1.58 x 10^-7 mol
2. let x moles of NaOH is added,
Buffer A,
pH = 5.10
[H3O+] = 7.91 x 10^-6 M
new pH = 4.10
new [H3O+] = 7.91 x 10^-5 M
moles of NaOH to be added = (7.91 x 10^-5 - 7.91 x 10^-6) x 0.02 L
= 1.42 x 10^-6 mol
3. Buffer B with greater concentration of NaC2H3O2 and HC2H3O2 has higher buffer capacity as it resists pH change to a wider range due to addition of acid or base to the system as compared to low concentration of Buffer A.
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