- #1

jackthehat

- 41

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Hi,

I have a problem which I have attempted to solve several times without success and i am at a loss to know where I am going wrong. The problem is ..

If 10.0 g of potassium sulfite is dissolved in water to make 200.0 mL of solution, then what is the molar concentration of potassium ions in the solution? (Hint: You should write a balanced chemical equation for the dissolution of potassium sulfite in water.)

My attempt ..

I did the thing in 3 steps ..

Step 1 - Find the Molarity of the solute ..

First find formula of compound .. compound comprises 2 elements 'K' (1+ ionisation number) and 'SO3' (2- ionisation number) so formula for compound is ... K2SO3 .. Molar mass of K2SO3 is ((2x39.098)+(1x32.06)+(3x15.999)) g/mol = 158.253 g/mol.

So Moles of (K2SO3) = (10.0 g/158.253) mol = 0.0632 mol

Now we have to account of the volume of the solution ..

Moles of solute (in solution) = 0.0632/100ml x 1L/1000ml = (0.0632 / 0.200) mol = 0.31595 mol

Step 2 - Find 'ion-to-solute' ratio

Reaction equation - K2SO3(s) --> 2K 1+ + SO3 2-

From the above reaction equation we see 2 moles of K1+ are produced from 1 mole of K2SO3 .. ratio is 2moles (K1+)/1 mole (K2SO3) =2/1

Ste 3 - Find ion molarity

Molarity of K1+ ion = Molarity of K2SO3 x (2/1) = 0.3159 x 2 = 0.63190 M .. ~ 0.632M (to 3 significant figures).

Am I attempting this all wrong or am I doing it the right way .. can you help please ?

Regards

Jack

I have a problem which I have attempted to solve several times without success and i am at a loss to know where I am going wrong. The problem is ..

If 10.0 g of potassium sulfite is dissolved in water to make 200.0 mL of solution, then what is the molar concentration of potassium ions in the solution? (Hint: You should write a balanced chemical equation for the dissolution of potassium sulfite in water.)

My attempt ..

I did the thing in 3 steps ..

Step 1 - Find the Molarity of the solute ..

First find formula of compound .. compound comprises 2 elements 'K' (1+ ionisation number) and 'SO3' (2- ionisation number) so formula for compound is ... K2SO3 .. Molar mass of K2SO3 is ((2x39.098)+(1x32.06)+(3x15.999)) g/mol = 158.253 g/mol.

So Moles of (K2SO3) = (10.0 g/158.253) mol = 0.0632 mol

Now we have to account of the volume of the solution ..

Moles of solute (in solution) = 0.0632/100ml x 1L/1000ml = (0.0632 / 0.200) mol = 0.31595 mol

Step 2 - Find 'ion-to-solute' ratio

Reaction equation - K2SO3(s) --> 2K 1+ + SO3 2-

From the above reaction equation we see 2 moles of K1+ are produced from 1 mole of K2SO3 .. ratio is 2moles (K1+)/1 mole (K2SO3) =2/1

Ste 3 - Find ion molarity

Molarity of K1+ ion = Molarity of K2SO3 x (2/1) = 0.3159 x 2 = 0.63190 M .. ~ 0.632M (to 3 significant figures).

Am I attempting this all wrong or am I doing it the right way .. can you help please ?

Regards

Jack

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