A stock solution contains 51.24 g of barium hydroxide in 1.20 L of solution. How do I make 1.00 L of 0.100 mol/L barium hydroxide from the stock solution?

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Answer 1 · 2014-09-15T01:37:23

You must dilute 401 mL with enough water to make 1.00 L of solution.

Step 1. Calculate the molarity of the stock solution.

Moles of Ba(OH)₂ = $"51.24 g Ba(OH)"_2 × ("1 mol Ba(OH)"_2)/("171.34 g Ba(OH)"_2) = "0.299 05 mol Ba(OH)"_2$

Molarity = $"moles"/"litres" = "0.299 05 mol"/"1.20 L"$ = 0.2492 mol/L

Step 2. Calculate the volume of stock solution that is needed.

$c_1V_1 = c_2V_2$

$c_1$ = 0.2492 mol/L; $V_1$ = ?
$c_2$ = 0.100 mol/L; $V_2$ = 1.00 L

$V_1 = V_2 × c_2/c_1$ = $"1.00 L" × "0.100 mol/L"/"0.2492 mol/L"$ = 0.401 L = 401 mL

So, to make your solution, you would dilute 401 mL of stock solution with enough water to make 1.00 L.