Question

HELP ASAP if you're good at gen chem I /AP chem: A student wishes to prepare a solution with a final concentration of Na+ = 0.50 M and a final concentration of HCO3– = 0.10 M by taking some NaOH and some trona (Na2CO3•NaHCO3•2H2O) and diluting with water to a final volume of 1.00 L. How much NaOH and trona are required? the answer is 0.05 mol NaOH and 0.15 mol trona, but I just don't understand how in the world to get there!?

Answer 1

Moles of trona are 0.15moles

Moles of NaOH are 0.05mol

Step-by-step explanation:

In the problem, you have 2 solutes, NaOH and trona.

Based in its formula, in solution, 1 mole of NaOH will produce 1 mole of Na⁺ and 1 mole of trona will produce 3 moles of Na⁺ and 1 mole of HCO₃⁻.

Also, you must know HCO₃⁻ will react with NaOH producing CO₃²⁻.

You want 1L of 0.50M Na⁺ = 0.50 moles of Na⁺ and 1L of HCO₃⁻ = 0.10M.

Now, in equations, moles of Na⁺ will be:

Moles Na⁺ = 0.50moles = X + 3Y (1)

Where X are moles of NaOH and Y moles of trona.

And, moles of HCO₃⁻ will be:

Moles HCO₃⁻ = 0.10moles = Y - X (2)

The sum of (1) + (2):

0.60mol = 4Y

0.15mol = Y

Moles of trona are 0.15moles

And moles of NaOH are (Replacing in (1) or (2) you will obtain the same, I will replace in (1)) :

0.50moles = X + 3Y

0.50moles = X + 3×0.15mol

0.50moles = X + 0.45mol

0.05 moles = X

Moles of NaOH are 0.05mol