Which of the following statements are FALSE? ⅰ Multiple answers: Multiple answers are accepted for this question Select one or more answers and submit. For keyboard navigation… SHOW MORE a If the pka for formic acid is 3.75 the ratio of formate (A) to formic acid (HA) at pH 4.15 is 5:1. the Ka = 0.00001. None b The pH of 1 liter of water to which is added 10 mL of 5.0 M HCI is 1.3. Which of the following C If an acid has a pKa of 4.76 the Ka = 0.00001. d The pH of 1 liter of water to which is added 10 ml of 0.50 M NaOH is 11.0. e The pH of a 1 liter solution to which has been added 6.0 mL of 1.5 M acetic acid and 5.0 mL of 0.4 M sodium acetate is 6.0.
The Correct Answer and Explanation is :
Let’s evaluate each statement for its truth or falsity:
a) If the pKa for formic acid is 3.75, the ratio of formate (A) to formic acid (HA) at pH 4.15 is 5:1. The Ka = 0.00001.
This statement is false. We can calculate the ratio of formate (A) to formic acid (HA) using the Henderson-Hasselbalch equation:
[
\text{pH} = \text{pKa} + \log \left( \frac{[A^-]}{[HA]} \right)
]
Substituting the given values:
[
4.15 = 3.75 + \log \left( \frac{[A^-]}{[HA]} \right)
]
[
0.4 = \log \left( \frac{[A^-]}{[HA]} \right)
]
[
\frac{[A^-]}{[HA]} = 10^{0.4} \approx 2.51
]
The ratio is about 2.51:1, not 5:1. Therefore, this statement is false.
b) The pH of 1 liter of water to which is added 10 mL of 5.0 M HCl is 1.3.
This statement is true. To calculate the pH, first, determine the moles of HCl added:
[
\text{moles of HCl} = 5.0 \, M \times 0.01 \, L = 0.05 \, \text{moles}
]
Now, calculate the concentration of HCl in the final volume (1 L of water + 0.01 L of HCl = 1.01 L):
[
\text{[HCl]} = \frac{0.05 \, \text{moles}}{1.01 \, \text{L}} \approx 0.0495 \, \text{M}
]
pH is the negative logarithm of the hydrogen ion concentration:
[
\text{pH} = -\log(0.0495) \approx 1.3
]
Thus, the statement is true.
c) If an acid has a pKa of 4.76, the Ka = 0.00001.
This statement is true. Ka is the acid dissociation constant, and it is related to pKa by the equation:
[
\text{pKa} = -\log(\text{Ka})
]
If pKa = 4.76:
[
\text{Ka} = 10^{-4.76} \approx 1.7 \times 10^{-5}
]
So, this statement is true as the Ka value is approximately (0.00001).
d) The pH of 1 liter of water to which is added 10 mL of 0.50 M NaOH is 11.0.
This statement is true. First, calculate the moles of NaOH added:
[
\text{moles of NaOH} = 0.50 \, M \times 0.01 \, L = 0.005 \, \text{moles}
]
Now, the concentration of OH⁻ in the final volume (1 L of water + 0.01 L of NaOH = 1.01 L):
[
\text{[OH}^-] = \frac{0.005 \, \text{moles}}{1.01 \, \text{L}} \approx 0.00495 \, \text{M}
]
To find the pH, we first find the pOH:
[
\text{pOH} = -\log(0.00495) \approx 2.31
]
Then, using the relationship ( \text{pH} + \text{pOH} = 14 ):
[
\text{pH} = 14 – 2.31 = 11.0
]
Thus, this statement is true.
e) The pH of a 1 liter solution to which has been added 6.0 mL of 1.5 M acetic acid and 5.0 mL of 0.4 M sodium acetate is 6.0.
This statement is false. We need to use the Henderson-Hasselbalch equation again, but first, we need to calculate the moles of acetic acid and acetate:
- Moles of acetic acid: (1.5 \, M \times 0.006 \, L = 0.009 \, \text{moles})
- Moles of sodium acetate: (0.4 \, M \times 0.005 \, L = 0.002 \, \text{moles})
Now, calculate the concentrations in the final volume (1 L + 0.006 L + 0.005 L = 1.011 L):
- Concentration of acetic acid: ( \frac{0.009}{1.011} \approx 0.0089 \, M )
- Concentration of acetate: ( \frac{0.002}{1.011} \approx 0.0020 \, M )
Now, apply the Henderson-Hasselbalch equation:
[
\text{pH} = \text{pKa} + \log \left( \frac{[A^-]}{[HA]} \right)
]
For acetic acid, pKa ≈ 4.76:
[
\text{pH} = 4.76 + \log \left( \frac{0.0020}{0.0089} \right) \approx 4.76 + \log(0.225) \approx 4.76 – 0.653 = 4.11
]
Thus, the pH is approximately 4.11, not 6.0. Therefore, this statement is false.
Summary of answers:
- a: False
- b: True
- c: True
- d: True
- e: False