1) A saturated solution of the slightly soluble salt MA3 is 1.4 ×10^–3 M in MA3. What is its Ksp value?
2) A 20.0 mL sample of NaIO3 was treated with an excess of KI and HCl and the iodine produced titrated with 0.05125 M Na2S2O3. 19.35 mL of Na2S2O3 solution were required to reach a starch‐iodine endpoint. What was the concentration of the NaIO3?
1) Ksp? More like K(lown)s(p) - the constant where the salt MA3 decides to get a little wild and go party in the solution! So, to find the Ksp value, we need to look at the concentration of MA3 in the saturated solution. In this case, it's 1.4 ×10^–3 M.
Now, Ksp is all about equilibrium, and the equation looks a bit like MA3 (s) ↔ M+ (aq) + 3A- (aq). Since we have 1 mole of MA3 producing 1 mole of M+ and 3 moles of A-, we can write the Ksp equation as:
Ksp = [M+]·[A-]^3.
Plugging in the concentration values, we get Ksp = (1.4 ×10^–3)·(1.4 ×10^–3)^3. So put on your math goggles and calculate that Ksp!
2) Oh, titration, we meet again! So we have a 20.0 mL sample of NaIO3, and we're adding excess KI and HCl, just to spice things up. Then we titrate the iodine produced with Na2S2O3, like a referee stepping in to calm things down.
We know that 19.35 mL of the Na2S2O3 solution were required to reach the starch-iodine endpoint. But hold your horses, because we're not done yet! The reaction equation between I2 (from NaIO3) and Na2S2O3 is:
I2 + 2S2O3^2- ↔ 2I- + S4O6^2-.
The mole ratio is 1:2, so if we used 19.35 mL of 0.05125 M Na2S2O3, that means we reacted with:
19.35 mL * 0.05125 M Na2S2O3 * (2 moles S2O3^2- / 1 mole I2).
Now, multiply that by the dilution factor, since the original sample was 20.0 mL. Divide that by 1000 to get the concentration in moles per liter. Finally, we can get the concentration of NaIO3:
Concentration of NaIO3 = (19.35 mL * 0.05125 M * 2) / (20.0 mL * 1000).
Now let's put on our lab coats and calculate that concentration!
1) To find the Ksp value, we need to know the concentration of the saturated solution and the stoichiometry of the salt solubility equation.
From the question, we know the saturated solution of MA3 has a concentration of 1.4 × 10^-3 M.
The stoichiometry of the salt solubility equation is as follows: MA3 (s) ↔ M^+(aq) + 3A^-(aq)
Since MA3 dissociates into one cation (M^+) and three anions (A^-), the solubility product expression can be written as follows:
Ksp = [M^+][A^-]^3
Considering that the concentration of each ion is the same as the concentration of the saturated solution, we can substitute the value and calculate the Ksp value:
Ksp = (1.4 x 10^-3)(1.4 x 10^-3)^3
Ksp = 2.74496 x 10^-12
Therefore, the Ksp value of the slightly soluble salt MA3 is 2.74496 x 10^-12.
2) In this problem, we will use the principle of stoichiometry to determine the concentration of NaIO3.
The balanced equation for the reaction is:
5 NaIO3 + 5 KI + 6 HCl -> 5 NaCl + 5 I2 + 6 KCl + 3 H2O
First, let's calculate the number of moles of Na2S2O3 used in the titration:
Moles of Na2S2O3 = Volume of Na2S2O3 solution (L) x Concentration of Na2S2O3(M)
= 0.01935 L x 0.05125 mol/L
= 9.926 x 10^-4 mol
According to the stoichiometry of the reaction, 1 mole of I2 is equivalent to 2 moles of Na2S2O3. Therefore, the number of moles of I2 produced in the reaction is:
Moles of I2 = 0.5 x Moles of Na2S2O3
= 0.5 x 9.926 x 10^-4 mol
= 4.963 x 10^-4 mol
From the balanced equation, we know that 1 mole of NaIO3 produces 1 mole of I2. Therefore, the number of moles of NaIO3 in the solution is also:
Moles of NaIO3 = Moles of I2
= 4.963 x 10^-4 mol
Finally, to find the concentration of NaIO3, divide the number of moles by the volume of the sample:
Concentration of NaIO3 = Moles of NaIO3 / Volume of Sample (L)
= 4.963 x 10^-4 mol / 0.020 L
= 2.4815 x 10^-2 M
Therefore, the concentration of NaIO3 in the sample is 2.4815 x 10^-2 M.
1) To find the Ksp value of a slightly soluble salt, you can use the information provided about its saturated solution concentration.
Ksp is the solubility product constant, which represents the equilibrium constant for a solid dissolving in a solution. It is defined as the product of the concentration of the ions raised to the power of their stoichiometric coefficients in the balanced chemical equation for the dissolution reaction.
For the salt MA3, let's assume its dissolution reaction is:
MA3(s) ⇌ 3M+(aq) + A-(aq)
The Ksp expression for this reaction would be:
Ksp = [M+]^3 * [A-]
Given that the saturated solution of MA3 is 1.4 × 10^–3 M in MA3, we know that the equilibrium concentration of both M+ and A- is 1.4 × 10^–3 M.
Substituting these values into the Ksp expression, we get:
Ksp = (1.4 × 10^–3)^3 * (1.4 × 10^–3)
Simplifying this calculation, we find the Ksp value for MA3.
2) To determine the concentration of NaIO3, we can use a titration reaction between NaIO3 and Na2S2O3.
The balanced chemical equation for this reaction can be written as follows:
2NaIO3 + 10Na2S2O3 + 12HCl → 10NaI + 12NaCl + 6S + 6H2O
From the equation, we see that 2 moles of NaIO3 react with 10 moles of Na2S2O3. This indicates a 1:5 stoichiometric ratio between the two compounds.
Given that 19.35 mL of 0.05125 M Na2S2O3 were required to reach the starch-iodine endpoint, we can use this information to determine the moles of Na2S2O3 used.
moles of Na2S2O3 = volume (in liters) * concentration
= 0.01935 L * 0.05125 mol/L
Since the stoichiometry of Na2S2O3 to NaIO3 is 5:1, we can determine the moles of NaIO3 used:
moles of NaIO3 = (moles of Na2S2O3) / 5
Knowing the moles of NaIO3 used, we can calculate its concentration in the 20.0 mL sample:
concentration of NaIO3 = moles / volume (in liters)
= (moles of NaIO3) / 0.020 L