Initial PH: 2.54 Malonic Acid (C3H4O4) Sodium Malonate ... - WizEdu
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- Initial pH: 2.54 malonic acid (C3H4O4) sodium malonate (NaC3H3O4) Addition of HCl (drop in pH) mL...
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In: Chemistry
Initial pH: 2.54 malonic acid (C3H4O4) sodium malonate (NaC3H3O4) Addition of HCl (drop in pH) mL...Initial pH: 2.54
malonic acid (C3H4O4) | sodium malonate (NaC3H3O4) |
Addition of HCl (drop in pH) | |
mL of HCl added | ΔpH |
4.12 | -0.51 |
6.93 | -0.69 |
11.59 | -0.89 |
25.63 | -1.09 |
49.9 | -1.24 |
Addition of NaOH (increase in pH) | |
mL of NaOH added | ΔpH |
1.91 | 2.77 |
2.61 | 3.00 |
3.56 | 3.21 |
4.43 | 3.48 |
4.98 | 3.69 |
5.39 | 3.96 |
5.87 | 4.25 |
6.36 | 4.51 |
6.9 | 4.71 |
Clearly outline how you made your assigned buffer with its target pH.
Assuming no volume change, how many mL of 0.1 M NaOH (use your concentration determined from standardization) should be added to your buffer for a pH change of 1.0?
concentration from standardization: .0945M HCl
Solutions
Expert Solution
Malonic acid is a diprotic acid having Ka1 = 1.4*10-3 ; therefore, pKa1 = -log (Ka1) = -log (1.4*10-3) = 2.85.
Use the Henderson-Hasslebach equation to find out the ratio of malonic acid and sodium malonate.
pH = pKa1 + log [NaC3H3O4]/[C3H4O4]
===> 2.54 = 2.85 + log [NaC3H3O4]/[C3H4O4]
===> -0.31 = log [NaC3H3O4]/[C3H4O4]
===> [NaC3H3O4]/[C3H4O4] = antilog (-0.31) = 0.4898 ≈ 0.49
Therefore, sodium malonate and malonic acid must be taken such that the ratio of their molar concentrations is 0.49:1. Since the total volume of the solution is not mentioned, neither is the total concentration of malonate given, hence, it is difficult to find out the individual concentrations of sodium malonate or malonic acid required to prepare the buffer (ans).
queen_honey_blossom answered 2 years ago Next > < PreviousRelated Solutions
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