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suter [353]
4 years ago
10

Which statement is correct? A. pKa is not an indicator of acid strength. B. An acid with a small Ka is stronger than an acid wit

h a large Ka. C. An acid with a large pKa is stronger than an acid with a small pKa. D. An acid with a small pKa is stronger than an acid with a large pKa.
Chemistry
2 answers:
Svetradugi [14.3K]4 years ago
3 0

Answer:

D

Explanation:

Since [pKa = - log Ka]....hence..,the larger the Ka value,the stronger the acid is..so this means that the pKa is vice versa

Saying that the smaller the pKa value..the stronger the acid is.

Fantom [35]4 years ago
3 0

Answer:

D) An acid with a small pKa is stronger than an acid with a large pKa

Explanation:

Hi, the acity constant (Ka) gives an idea of the strenght of an acid. The higher the Ka, the stronger the acid.

The indicator pKa is obtained by the following formula:

pKa=-\log{Ka}

Due to the logarithm, when the Ka increases, the pKa decreases.

<u>Putting all this toghether</u>: An acid with a large Ka has a small pKa and is stronger than an acid with a small Ka (large pKa).

So the correct statement is D

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Starting with 250 mL of a 0.250 M solution of HBr; a) Calculate the initial pH of the solution.b) Calculate the pH after adding
maxonik [38]

Answer:

a.

pH =  0.602

b.

pH = 1.5

c.

pH = 7

d.

pH = 12.1

Explanation:

a ) To calculate the pH, use the following equation:

pH = -log [H+]

Hbr is a strong acid, so the [H+] concentration can be calculated as follow:

[HBr] = 0.250 M

As acid Hbr:

Hbr = H+ + Br-

As strong acid HBr dissociates at all, so

[HBr] = [H+] = 0.250 M

So the pH:

<u>pH = -log [0.250 M] = 0.602</u>

<u></u>

<u>b) Calculate the pH after adding 250 mL of 0.125M NaOH</u>

<u></u>

<u>I</u>n this point, the reactions starts:

<u></u>

HBr + NaOH = H2O + NaBr

- First, we gonna find the mol of each reactant:

HBr:

mol = [M] × L

mol = 0.250 M × 0.250 L

mol = 0.0625 mol HBr

NaOH:

mol = [M] × L

mol = 0.125 M × 0.250L

mol = 0.03125 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.0625 mol HBr – 0.03125 mol HBr = 0.03125 mol HBr

These are the moles free in the solution, and we going to use them to calculate the pH:

pH = -log [ H]

pH = - log [ 0.03125 ] = 1.5

<u>c) Calculate the pH after adding 500 mL of 0.125M NaOH</u>

NaOH:

mol = [M] × L

mol = 0.125 M × 0.500L

mol = 0.0625 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.0625 mol HBr – 0.0625 mol HBr = 0 HBr

pH = 7

<u>d) Calculate the pH after adding 600 mL of 0.125M NaOH.</u>

<u>NaOH:</u>

mol = [M] × L

mol = 0.125 M × 0.600L

mol = 0.075 mol NaOH

In base on the reaction, it’s needed 1 mol of NaOH to neutralize 1 mole of HBr, so to neutralize 0.0625 moles of HBr its needed 0.0625 mol of NaOH:

0.075 mol NaOH – 0.0625 mol NaOH = 00125 NaOH

These are the moles free in the solution, and we going to use them to calculate the pH:

pOH = -log [ OH]

pOH = - log [ 0.0125 ] = 1.90

pH + pOH = 14

pH = 14- pOH = 14 – 1.90 = 12.1

4 0
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