Question

You titrate 25.00 mL of 0.03040 M Na2CO3 with 0.02790 M Ba(NO3)2. Calculate pBa2 after the following volumes of Ba(NO3)2 are added. Ksp for BaCO3 is 5.0 × 10–9. (a)12.50 ml (b) Veq (c) 34.70 ml

Answer 1

Reaction is

Ba(NO₃)₂ (aq) + Na₂CO₃ (aq) ---->BaCO₃(s) + 2 NaNO₃

a) 12.5 ml of Ba/NO3)2 , let´s calculate moles, moles = Molarity * volume

moles = 0.0279 * 0.0125 = 0.00034875 moles

moles of Na2CO3 = Molarity * volume = 0.0304 * 0.025 = 0.00076 moles

excess of Na2CO3 = 0.00076 - 0.00034875 = 0.00041125 moles

Molarity of Na2CO3 = 0.00041125 / (0.0375)=0.01096 M , Molarity = moles / volume

BaCO3 ===== Ba⁺² + CO₃

Ksp = [Ba] * [CO₃], multiplication of the concentrations

5 x 10^-9 = [Ba] * 0.01096

[Ba] = 5 x 10^-9 / 0.01096 = 4.56 x 10^-7 M

pBa = -log [Ba] = -log (4.56 x 10^-7) = 6.34

b) at equilibrium point

Ksp = [Ba] * [CO₃]

5 x10^-9 = x²

x = 5 x10^-9

x = 7.07107 x 10^-5

Pba = -log ( 7.07107 x 10^-5) = 4.15

c) calculate moles of every substance

moles of Na2CO3 = Molarity * volume = 0.0304 * 0.025 = 0.00076 moles

moles of Ba(NO3)2 = 0.0279 * 0.0347 = 0.00096813 moles

excess moles of Ba(NO3)2 = 0.00096813 - 0.00076 = 0.00020813 moles

total volume = 25 + 34.7 = 59.7 ml = 0.0597 L

Concentration of excess = 0.00020813 / 0.0597 = 0.003486 M

this will be also the concentration of Ba

Pba = -log(0.003486) = 2.45

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