1.
Calculate the pH of a buffer solution that contains 0.32 M benzoic acid (C6H5CO2H) and 0.17 M sodium benzoate (C6H5COONa). [Ka = 6.5 × 10-5 for benzoic acid] Round your answer to two places past the decimal.
2.
A solution is prepared by mixing 470 mL of 0.18 M Tris·Base and 598 mL of 0.27 M Tris·Acid. What is the pH of this solution? [Ka(Tris·Acid) = 8.5 × 10-9] Round your answer to two places past the decimal.
3.
Calculate the pH of a buffer solution prepared by dissolving 0.21 mole of 3-chloropropanoic acid, HC3H4O2Cl and 0.86 mole of sodium 3-chloropropanoate, NaC3H4O2Cl in enough water to make 0.79 liter of solution. [Ka(HC3H4O2Cl) = 7.8 × 10-5] Round your answer to two places past the decimal.
4.
You are asked to go into the lab and prepare an acetic acid-sodium acetate buffer solution with a pH of 5.08 ± 0.02. What molar ratio of CH3COONa (conjugate base) to CH3COOH (weak acid) should be used? Read carefully. Round your answer to two places past the decimal.
5.
A buffer is prepared by adding 0.91 L of 0.99 M HCl to 815 mL of 1.5 M NaHCOO. What is the pH of this buffer? [Ka(HCOOH) = 1.7 × 10-4] Round your answer to two places past the decimal.
6.
A certain monoprotic weak acid with Ka = 0.44 can be used in various industrial processes. (a) What is the [H+] for a 0.202 M aqueous solution of this acid and (b) what is its pH? Round the [H+] to three significant figures and the pH to two places past the decimal and do NOT use scientific notation.
7.
What is the pH of 65 mL of a 1.3 M benzoic acid (HC7H5O2, Ka=6.4x10–5) solution after 20 mL of a 0.33 NaOH solution is added? Round your answer to two places past the decimal.
#1. = 3.91
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1. To calculate the pH of a buffer solution, we need to use the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log ([A-] / [HA])
where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
In this case, the weak acid is benzoic acid (C6H5CO2H) and the conjugate base is sodium benzoate (C6H5COONa). The Ka value for benzoic acid is 6.5 × 10-5.
First, we need to calculate the ratio [A-] / [HA] using the molar concentrations of sodium benzoate and benzoic acid. Given that the concentration of benzoic acid is 0.32 M and the concentration of sodium benzoate is 0.17 M, we can substitute these values into the equation:
[A-] / [HA] = 0.17 / 0.32
Next, we calculate the pKa by taking the negative logarithm of the Ka value:
pKa = -log(6.5 × 10-5)
Finally, we substitute the calculated values into the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log ([A-] / [HA])
Note: be sure to take the logarithm with the appropriate base depending on the calculator or software being used.
2. To calculate the pH of the solution, we can use the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log ([A-] / [HA])
where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
In this case, we have Tris·Base and Tris·Acid. The Ka value for Tris·Acid is 8.5 × 10-9.
First, we need to calculate the ratio [A-] / [HA] using the molar concentrations of Tris·Base and Tris·Acid. Given that the concentration of Tris·Base is 0.18 M and the concentration of Tris·Acid is 0.27 M, we can substitute these values into the equation:
[A-] / [HA] = 0.27 / 0.18
Next, we calculate the pKa by taking the negative logarithm of the Ka value:
pKa = -log(8.5 × 10-9)
Finally, we substitute the calculated values into the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log ([A-] / [HA])
Note: be sure to take the logarithm with the appropriate base depending on the calculator or software being used.
3. To calculate the pH of the buffer solution, we can use the Henderson-Hasselbalch equation, which is given by:
pH = pKa + log ([A-] / [HA])
where pKa is the negative logarithm of the acid dissociation constant (Ka), [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.
In this case, we have 3-chloropropanoic acid (HC3H4O2Cl) and sodium 3-chloropropanoate (NaC3H4O2Cl). The Ka value for HC3H4O2Cl is 7.8 × 10-5.
First, we need to calculate the ratio [A-] / [HA] using the molar concentrations of NaC3H4O2Cl and HC3H4O2Cl. Given that the moles of NaC3H4O2Cl is 0.86 and the moles of HC3H4O2Cl is 0.21, we can calculate their molar concentrations:
[A-] = 0.86 moles / 0.79 L
[HA] = 0.21 moles / 0.79 L
Next, we calculate the pKa by taking the negative logarithm of the Ka value:
pKa = -log(7.8 × 10-5)
Finally, we substitute the calculated values into the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log ([A-] / [HA])
Note: be sure to take the logarithm with the appropriate base depending on the calculator or software being used.
4. To determine the molar ratio of CH3COONa (conjugate base) to CH3COOH (weak acid) in the acetic acid-sodium acetate buffer solution, we can use the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-] / [HA])
In this case, we are given a desired pH of 5.08 ± 0.02 and the pKa of acetic acid (CH3COOH) is known to be 4.76.
First, we subtract the pKa from the desired pH to find the log ratio:
log ([A-] / [HA]) = pH - pKa
Next, we take the antilog of both sides to find the ratio:
([A-] / [HA]) = 10^(pH - pKa)
Finally, we can determine the molar ratio of CH3COONa to CH3COOH by dividing the concentration of the conjugate base ([A-]) by the concentration of the weak acid ([HA]).
5. To calculate the pH of the buffer, we need to use the Henderson-Hasselbalch equation. Given the concentrations of HCl and NaHCOO, and the Ka value for HCOOH, we can calculate the pH as follows:
First, we need to determine the moles of HCl and NaHCOO:
Moles of HCl = 0.91 L x 0.99 M
Moles of NaHCOO = 0.815 L x 1.5 M
Next, we calculate the concentration of the weak acid (HCOOH) and the conjugate base (HCOO-):
[HCOOH] = Moles of NaHCOO / Total volume of the solution
[HCOO-] = Moles of HCl / Total volume of the solution
Then, we use the Henderson-Hasselbalch equation:
pH = pKa + log ([HCOO-] / [HCOOH])
Substitute the calculated values into the equation and solve for pH.
6. (a) To calculate [H+], we need to take the square root of the Ka value. In this case, the Ka value is 0.44, so:
[H+] = sqrt(Ka)
Note: be sure to use the square root function properly when using a calculator or software.
(b) To calculate the pH, we use the equation:
pH = -log[H+]
Substitute the calculated value of [H+] to find the pH.
7. To calculate the pH of the solution after adding NaOH, we need to determine the overall reaction between benzoic acid (HC7H5O2) and NaOH.
The balanced chemical equation for the reaction is:
HC7H5O2 + OH- → C7H5O2- + H2O
First, calculate the moles of benzoic acid in the initial solution:
Moles of HC7H5O2 = 65 mL x 1.3 M
Next, calculate the moles of NaOH added:
Moles of NaOH = 20 mL x 0.33 M
Since NaOH is in excess, we have a net reaction that consumes all of the benzoic acid and produces the benzoate ion (C7H5O2-).
Calculate the moles of benzoate ion formed:
Moles of C7H5O2- = Moles of HC7H5O2
Finally, calculate the concentration of the benzoate ion and use it to determine the pH using the Henderson-Hasselbalch equation:
[H+] = sqrt(Ka * [C7H5O2-] / [HC7H5O2])
pH = -log[H+]
Substitute the calculated values into these equations to find the pH.