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I tried these four problems. I am giving my answers. What did I do wrong? Could

ID: 826856 • Letter: I

Question

I tried these four problems. I am giving my answers. What did I do wrong? Could someone help.

I will award points quickly. I got behind in class due to hospital stay from accident

For number 2 and 3 I used pka + log (base/acid) to get answer



If the pKa is 1.689 what is the value of Ka? (Note: do not use scientific notation)


Answer Given: 0.020


I used Ka= 10^-pKa


Find the pH of a solution that has a weak acid concentration of 0.74 M and a weak base concentration of 0.85 M, and a pKa of 4.63 .


Answer Given: 4.77


If 10.90 mL of an Ammonia solution required 10.48 mL of 0.9723 M HCl to reach the endpoint, what's the concentration of the Ammonia solution?


Answer Given: 5.0


Find the pH of a solution that has a weak acid concentration of 1.20 M and a weak base concentration of 1.12 M, and a pKa of 6.89 .


Answer Given: 6.82

Explanation / Answer

1). If the pKa is 1.689 what is the value of Ka?

We know pKa is simply the -log(Ka), so the easiest way is to solve as shown:

pKa = -log(Ka)

1.689 = -log(Ka)

-1.689 = log(Ka)

10^-1.689 = Ka

Ka = 0.0204


2).Find the pH of a solution that has a weak acid concentration of 0.74 M and a weak base concentration of 0.85 M, and a pKa of 4.63.

Use the henderson-hasselbach equation:

pH = pKa + log(base/acid)

pH = 4.63 + log(0.85/0.74)

pH = 4.69


3). If 10.90 mL of an Ammonia solution required 10.48 mL of 0.9723 M HCl to reach the endpoint, what's the concentration of the Ammonia solution?

Use the equation M1*V1 = M2*V2

Let 1 be the ammonia and 2 be the HCl

M1*V1 = M2*V2

M1*10.9 = 0.9723*10.48

M1 = (0.9723*10.48)/10.9

M1 = 0.9348

This works because we are titrating a strong base with a strong acid, so they fully dissociate


4). Find the pH of a solution that has a weak acid concentration of 1.20 M and a weak base concentration of 1.12 M, and a pKa of 6.89 .

Here we use the henderson-hasselbach equation again.

pH = pKa + log(base/acid)

pH = 6.89 + log(1.12/1.2)

pH = 6.86


Most of your answers were close, but there may have been a calculation error