Now, consider the beaker containing 1.00 L of a solution of equal concentrations
ID: 580603 • Letter: N
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. 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 H PO4 and HPO4? 5. 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 HPO4? 6.Explanation / Answer
Q5
in beaker C:
1 M of Hcl = 1 M of H+
pH = -log(H+) = -log(1)
pH = 0
when it is added to H2PO4- and HPO4-2 then, H+ decreases HPO4-
mmol of HPO4-2 = MV = 1*0.01 = 0.01 mol = 10 mmol
mmol of H2PO4- = MV = 1*0.01 = 0.01 mol = 10 mmol
after addition of H+
mmol ofH+ = MV = 1*1 = 1
mmol of HPO4-2 = 10-1
mmol of H2PO4- = 10+1
pH = pKa + log(HPO4-2 / H2PO4-)
pH = 7.21 + log(9/10) = 7.164
initial pH = 7.21, since log(10/10) = 1
Q6
similarly
[OH-] = 1
pOH = -log(OH) = -log(1) = 0
pH = 14
when it is added to H2PO4- and HPO4-2 then, H+ decreases HPO4-
mmol of HPO4-2 = MV = 1*0.01 = 0.01 mol = 10 mmol
mmol of H2PO4- = MV = 1*0.01 = 0.01 mol = 10 mmol
after addition of OH-
mmol of OH- = MV = 1*1 = 1
mmol of HPO4-2 = 10+1
mmol of H2PO4- = 10-1
pH = pKa + log(HPO4-2 / H2PO4-)
pH = 7.21 + log(10/9) = 7.255
initial pH = 7.21, since log(10/10) = 1
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