Question

Determine the molarity of supplied Na2CO3 solution by titrating against
M
H2SO4 solution.
20​

Answer 1

Explanation:

Here is what I did and got the answer and don't understand why I need to do it:

1 g Na2CO3⋅10H2O286 g Na2CO3⋅10H2O0.003447 mol0.020 L(0.174825 M)(0.020 L)x=0.003447 mol=0.174825 M=(0.250 L)(x)=0.013986 M

1 g NaX2COX3⋅10HX2O286 g NaX2COX3⋅10HX2O=0.003447 mol0.003447 mol0.020 L=0.174825 M(0.174825 M)(0.020 L)=(0.250 L)(x)x=0.013986 M

0.013986 M0.013986 M is the correct answer, but I don't know why and don't understand the question. If you were to do the problem, could you explain why you did what you did to get to the answer

Answer 2

The molarity of the supplied Na2CO3 solution is 0.125.

To determine the molarity of the Na2CO3 solution by titrating against H2SO4, we need to perform an acid-base titration. The balanced chemical equation for the reaction between Na2CO3 and H2SO4 is:

H2SO4 plus Na2CO3 = H2SO4 plus H2O + CO2

From this equation, we can see that 1 mole of H2SO4 reacts with 1 mole of Na2CO3.

To perform the titration, we first need to measure a known volume of the Na2CO3 solution using a volumetric pipette and transfer it to a clean Erlenmeyer flask. Then, we add a few drops of phenolphthalein indicator to the solution. The solution will turn pink if the pH is above 8.2.

Next, we titrate the Na2CO3 solution with the H2SO4 solution by adding it slowly to the Na2CO3 solution while stirring until the pink color of the indicator disappears. This indicates that all the Na2CO3 has reacted with the H2SO4.

At this point, we can record the volume of H2SO4 used in the titration. Let's say we used 25.0 mL of 0.1 M H2SO4.

Now, we can use the equation M1V1 = M2V2 to calculate the molarity of the Na2CO3 solution. M1 is the molarity of the H2SO4 solution, V1 is the volume of the H2SO4 solution used in the titration, M2 is the molarity of the Na2CO3 solution, and V2 is the volume of the Na2CO3 solution used in the titration.

We know M1 is 0.1 M, V1 is 25.0 mL, and we assume that the volume of Na2CO3 solution used in the titration is 20.0 mL.To solve for M2, we can rewrite the equation as follows:

M2 = (M1V1) / V2

M2 = (0.1 M) x (25.0 mL) / (20.0 mL)

M2 = 0.125 M

Therefore, the molarity of the supplied Na2CO3 solution is 0.125 M.

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