Each of your test solutions contained 10.00 ml. of 0.100 M NaA and 10.00 ml of 0

Question

Each of your test solutions contained 10.00 ml. of 0.100 M NaA and 10.00 ml of 0.100 M HA. Keeping in mind the equations you wrote above, consider by answering llowing questions what would have happened to the pH if instead of adding 2.00 How many mmols of A are present initially? How many mmols of HA are present initially? How many mmol of H are added from the more concentrated acid (HNO,)? mL of 0.01M HCI to solution G you had added 6.50 mL of 0.45 M HNO. Show your calculations. 0,mLxO.lo.mnel Which component of solution G reacts with the added HNO,? Write the chemical equation for the reaction: How many mmol of the added HNO: remain after reaction with solution G? What is [H'] in the final solution? What is the expected pH of this solution? Can this solution (G+ HNOs) still be classified as a buffer? Why or why not?

Explanation / Answer

Molarity = Moles / Liter

Moles = Molarity X Liter

mMoles = Molarity x milliliter

Now

10.00 mL of 0.100 M NaA

So, mmoles of A- present = 0.100 M x 10 mL = 1.00 mmol

10.00 mL of 0.100 M HA

mmoles of HA present = 0.100 M x 10 mL = 1.00 mmol

6.50 mL of 0.45 M HNO3

mmoles of HNO3 = 0.45 M x 6.50 mL = 2.925 mmol

HNO3 is a strong acid and it will dissociate completely

So, mmoles of H+ = 2.925 mmol

NaA reacts with HNO3 to form HA.

A-      +     H+     =     HA

mmoles of A- present = 1.00 mmol

mmoles of HNO3 (H+ ) = 2.925 mmol

So, 1.00 mmoles of A- reacts with 1.00 mmoles of HNO3 (H+ ).

So, mmoles of HNO3 remain after reaction = 2.925 – 1.00 = 1.925 mmoles

In final solution you have 1.925 mmoles of HNO3 (or H+ )

Total Volume = 10.00 mL + 10.00 mL + 6.50 mL = 26.50 mL

So, [H+] = 1.925 mmol / 26.50 mL = 0.073 M

pH = -log[H+] = -log(0.073) = 1.14

C+ HNO3 will not be a buffer. Since a buffer is a mixture of a weak acid and its conjugated base and vice versa. But in this case after adding HNO3, all the conjugated base (NaA) is neutralized and only remains HNO3 which is a strong acid.

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