{"id":182573,"date":"2025-01-14T11:48:04","date_gmt":"2025-01-14T11:48:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182573"},"modified":"2025-01-14T11:48:06","modified_gmt":"2025-01-14T11:48:06","slug":"which-of-the-following-statements-are-false","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/14\/which-of-the-following-statements-are-false\/","title":{"rendered":"Which of the following statements are FALSE"},"content":{"rendered":"\n

Which of the following statements are FALSE? \u2170 Multiple answers: Multiple answers are accepted for this question Select one or more answers and submit. For keyboard navigation\u2026 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.<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Let’s evaluate each statement for its truth or falsity:<\/p>\n\n\n\n

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.<\/strong>
This statement is false<\/strong>. We can calculate the ratio of formate (A) to formic acid (HA) using the Henderson-Hasselbalch equation:<\/p>\n\n\n\n

[
\\text{pH} = \\text{pKa} + \\log \\left( \\frac{[A^-]}{[HA]} \\right)
]<\/p>\n\n\n\n

Substituting the given values:<\/p>\n\n\n\n

[
4.15 = 3.75 + \\log \\left( \\frac{[A^-]}{[HA]} \\right)
]<\/p>\n\n\n\n

[
0.4 = \\log \\left( \\frac{[A^-]}{[HA]} \\right)
]<\/p>\n\n\n\n

[
\\frac{[A^-]}{[HA]} = 10^{0.4} \\approx 2.51
]<\/p>\n\n\n\n

The ratio is about 2.51:1, not 5:1. Therefore, this statement is false.<\/p>\n\n\n\n

b) The pH of 1 liter of water to which is added 10 mL of 5.0 M HCl is 1.3.<\/strong>
This statement is true<\/strong>. To calculate the pH, first, determine the moles of HCl added:<\/p>\n\n\n\n

[
\\text{moles of HCl} = 5.0 \\, M \\times 0.01 \\, L = 0.05 \\, \\text{moles}
]<\/p>\n\n\n\n

Now, calculate the concentration of HCl in the final volume (1 L of water + 0.01 L of HCl = 1.01 L):<\/p>\n\n\n\n

[
\\text{[HCl]} = \\frac{0.05 \\, \\text{moles}}{1.01 \\, \\text{L}} \\approx 0.0495 \\, \\text{M}
]<\/p>\n\n\n\n

pH is the negative logarithm of the hydrogen ion concentration:<\/p>\n\n\n\n

[
\\text{pH} = -\\log(0.0495) \\approx 1.3
]<\/p>\n\n\n\n

Thus, the statement is true.<\/p>\n\n\n\n

c) If an acid has a pKa of 4.76, the Ka = 0.00001.<\/strong>
This statement is true<\/strong>. Ka is the acid dissociation constant, and it is related to pKa by the equation:<\/p>\n\n\n\n

[
\\text{pKa} = -\\log(\\text{Ka})
]<\/p>\n\n\n\n

If pKa = 4.76:<\/p>\n\n\n\n

[
\\text{Ka} = 10^{-4.76} \\approx 1.7 \\times 10^{-5}
]<\/p>\n\n\n\n

So, this statement is true as the Ka value is approximately (0.00001).<\/p>\n\n\n\n

d) The pH of 1 liter of water to which is added 10 mL of 0.50 M NaOH is 11.0.<\/strong>
This statement is true<\/strong>. First, calculate the moles of NaOH added:<\/p>\n\n\n\n

[
\\text{moles of NaOH} = 0.50 \\, M \\times 0.01 \\, L = 0.005 \\, \\text{moles}
]<\/p>\n\n\n\n

Now, the concentration of OH\u207b in the final volume (1 L of water + 0.01 L of NaOH = 1.01 L):<\/p>\n\n\n\n

[
\\text{[OH}^-] = \\frac{0.005 \\, \\text{moles}}{1.01 \\, \\text{L}} \\approx 0.00495 \\, \\text{M}
]<\/p>\n\n\n\n

To find the pH, we first find the pOH:<\/p>\n\n\n\n

[
\\text{pOH} = -\\log(0.00495) \\approx 2.31
]<\/p>\n\n\n\n

Then, using the relationship ( \\text{pH} + \\text{pOH} = 14 ):<\/p>\n\n\n\n

[
\\text{pH} = 14 – 2.31 = 11.0
]<\/p>\n\n\n\n

Thus, this statement is true.<\/p>\n\n\n\n

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.<\/strong>
This statement is false<\/strong>. We need to use the Henderson-Hasselbalch equation again, but first, we need to calculate the moles of acetic acid and acetate:<\/p>\n\n\n\n