If 29.0 mL of 0.10 M Ba(NO3)2 are added to 76.0 mL of 0.21 M Na2CO3, will BaCO3 precipitate? Calculate Q
Q = (Ba^2+)(CO3^2-)
[Ba(NO3)2] = 0.10M x (29/105) = ?M
[Na2CO3] = 0.21 M x (76/105) = ?M
Multiply the concns and see how Q compares with Ksp.
To determine if BaCO3 will precipitate when Ba(NO3)2 is mixed with Na2CO3, we need to calculate the reaction quotient (Q). The reaction between Ba(NO3)2 and Na2CO3 can be represented as:
Ba(NO3)2 + Na2CO3 → BaCO3 + 2NaNO3
First, we need to identify the stoichiometric coefficients of the reactants and products. From the balanced equation, we can see that the ratio of moles of BaCO3 formed to Ba(NO3)2 is 1:1.
Now, let's calculate the moles of each compound in the reaction:
Moles of Ba(NO3)2 = volume (in L) × concentration (in mol/L)
= 29.0 mL × (1 L / 1000 mL) × 0.10 mol/L
= 0.0029 mol
Moles of Na2CO3 = volume (in L) × concentration (in mol/L)
= 76.0 mL × (1 L / 1000 mL) × 0.21 mol/L
= 0.016 mol
Now that we have the moles of Ba(NO3)2 and Na2CO3, we can calculate Q:
Q = [BaCO3] / [Ba(NO3)2]
= (moles of BaCO3 formed) / (moles of Ba(NO3)2)
Since the ratio of moles of BaCO3 formed to Ba(NO3)2 is 1:1, Q can be calculated as:
Q = 0.0029 mol / 0.0029 mol
= 1
When Q = 1, it means the reaction is in equilibrium. If Q is greater than 1, it means there is an excess of product present and precipitation occurs. However, if Q is less than 1, there is an excess of reactant remaining and no precipitation occurs.
In this case, Q is equal to 1, which means the reaction is at equilibrium. Therefore, BaCO3 will not precipitate.