Question

A 500 ml sample of 0.0040 M HCNO is titrated with 0.0020 M RbOH. Calculate the pH for the solution after the following additions of RbOH: 0 ml, 250 ml, 750 ml, 999 ml, 1000 ml, 1000 ml, 1001 ml, 1250 ml, 1500 ml.

Answer 1

moles of acids available = Molarity * Volume

moles = 0.004 * 0.5 = 0.002 moles of Acid

a) with 0 ml, this is the iniatl PH, since no base has been added the PH = -log(Conc. HCNO)

PH = - log (0.004) = 2.397

b) moles of base added = Molarity * Volume = 0.002 * 0.25 = 0.0005 moles

There are more moles of acid than base so : 0.002 - 0.0005 = 0.0015 moles,

total volume = 500 + 250 = 750 ml or 0.75 L

Molarity = 0.0015 / 0.75 = 0.002

PH = -log (0.002) = 2.69

c)

moles of base added = Molarity * Volume = 0.002 * 0.75 = 0.0015 moles

There are more moles of acid than base so : 0.002 - 0.0015 = 0.0005 moles,

total volume = 500 + 750 = 1250 ml or 1.25 L

Molarity = 0.0005 / 1.25 = 0.0004

PH = -log (0.0004) = 3.397

d) moles of base = Molarity * volume = 0.002 * 0.999 = 0.001998

moles of acid remaining = 0.002 -  0.001998 = 0.000002

total volume = 500 + 999 = 1499 ml or 1.499 L

Molarity = 0.000002 / 1.499 = 0.000001334

PH = 5.874

e) 1000 ml of Base

moles = 0.002 * 1 = 0.002

moles of acid = 0.002

this is the equivalence point, all the base will react with the acid so the PH will be equal to 7

d) Beyond the equivalence point we can expect to have OH ions on the solution so the PH will be higher than 7

1001 ml, moles = 0.002 * 1.001 = 0.002002

moles of acid = 0.002

there are more moles of base than acid so 0.002002 - 0.002 = 0.000002

total volume = 500 + 1001 = 1501 ml or 1.501

Molarity = 0.000002 / 1.501 = 0.000001332

POH (because we have ions of OH) = -log (0.000001332) = 5.87

PH + POH = 14

PH = 14 - 5.87 = 8.13

The rest follows the same pattern

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