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 the following questions what would have happened to the pH if instead of adding 2.00 mL of 0.01M HCI to solution G you had added 6.50 mL of 0.45 M HNO3. Show your calculations. How many mmols of A are present initially? 0100)(iG.oc) O 0t mna How many mmols of HA are present initially? 26. SC How many mmol of H' are added from the more concentrated acid (HNO:)? Which component of solution G reacts with the added HNOs? Write the chemical equation for the reaction: How many mmol of the added HNOs 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

initially

a)

mmol of A- = Msalt*Vsalt = (10)(0.1) = 1 mmol of A-

b)

mmol of HA = Macid*Vacid = 10*0.1 = 1 mmol of HA

c)

mmol of HNO3 = MHNO3 * VHNO3 = 6.5*0.45 = 2.925 mmol of HNO3 added

d)

from G

HNO3 is a strong acid, dissociates to form H+ and NO3-

H+ will react with A- to form HA

so choose A-

A-(aq) + H+(aq) --> HA(aq)

e)

mmol of HNO3 after reaction = 2.925 -1 = 1.925 mmol of HNO3 left

f)

[H+] = mmol of H+ / Total V = (1.925 ) /(10+10+6.5) = 0.07264 M

g)

pH = -log([H+]) = -log(0.07264 = 1.14

h)

not likely, since there is only HNO3 left, we require weak  acid + conjugate base only ( HA and A-)

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