Now, consider the beaker containing 1.00 L of a solution of equal concentrations

Question

Now, consider the beaker containing 1.00 L of a solution of equal concentrations (0.010 M) of dihydrogen phosphate and hydrogen phosphate to which either 1.0 mL of 1.0 M HCl (beaker C) or 1.0 mL of 1.0 M Na(OH) (beaker D) is added. a. Calculate the pH of the contents of Beaker C. How much does the pH change when a strong acid is added to an equal mixture of H2PO4 and HPO4? b. Calculate the pH of the contents of Beaker D. How much does the pH change when a strong base is added to an equal mixture of H2PO and HPo? C. What do you notice about the change in pH when an acid or base is added to pure water versus a solution containing equal amounts of mixture of a weak acid and its conjugate base? If there are differences between the two systems, propose a reason for the observed behavior.

Explanation / Answer

initial pH

pH = pKa + log(HPO4-2/H2Po4-)

pH = 7.21 + log(10/10)

pH = 7.21

a)

this is only acid so

pH = -log(H+) = -log(1) = 0

if we add it to the solution

mmol of H = MV = 1*1 = 1

mmol of H2PO4- = 1*0.01 = 0.01 mol = 10

mmol of HPO4-2 = 1*0.01 = 0.01 mol = 10

after reaction

mmol of H2PO4- =10+1 = 11

mmol of HPO4-2 = 10-1 = 9

pH = pKa + log(HPO4-2/H2PO4)

pH = 7.21 + log(9/11)=

7.122

dpH = 7.21 - 7.122 = 0.088 drop/decrease

b)

this is only base so

pOH = -log(OH-) = -log(1) = 0

if we add it to the solution

mmol of OH- = MV = 1*1 = 1

mmol of H2PO4- = 1*0.01 = 0.01 mol = 10

mmol of HPO4-2 = 1*0.01 = 0.01 mol = 10

after reaction

mmol of H2PO4- =10-1 = 9

mmol of HPO4-2 = 10+1 = 11

pH = pKa + log(HPO4-2/H2PO4)

pH = 7.21 + log(11/9)= 7.2971

dpH = 7.21 - 7.2971 = 0.0871 increase/rise

c)

when we add it to pure water --> pH changes drastically

when we add it to a buffer --> pH is buffered, it will not decrease/icnrease that mcuh

the system is essentially the equilbirium formation vs. non-equilbirium ( pure water)

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