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What volume of 0. 4 m ba (oh)2 must be added to 50 ml of 0. 3 m naoh to give a solution 0. 5 m in oh-

User Bidby
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Answer: 33 mL of 0.4 M Ba(OH2) must be added.

Step-by-step explanation:

We will solve this problem by tracking the number of moles in the solutions.

Consider the balanced chemical equation for the dissolution of Ba(OH)2:

Ba(OH)2 (aq) --> Ba^2+ (aq) + 2 OH- (aq)

From the coefficients we see that 2 moles of OH- ions are produced for every 1 mole of Ba(OH)2.

For NaOH:

NaOH (aq) --> Na^+ (aq) + OH- (aq),

so 1 mole of OH- ions are produced for every 1 mole of NaOH.

Number of OH- ions

From the information we gathered:

0.4 mol/L Ba(OH2) x 2 mol OH- / (1 mol Ba(OH)2) =0.8 mol/L OH-

0.3 mol/L NaOH x 1 mol OH- / (1 mol NaOH) = 0.3 mol/L OH-

We are asked to find the volume, X, required to give a solution of 0.5 M in OH-.

(0.8 mol/L of OH-) x (X) = 0.8X mol OH-

(0.3 mol/L of OH-) x (0.050 L) = 0.015 mol OH-

Our goal is:

(0.5 mol/L of OH-) x (X + 0.050 L) = 0.5(X + 0.050) mol OH-

Setting the number of moles equal:

0.8X + 0.015 = 0.5(X + 0.050)

0.8X + 0.015 = 0.5X + 0.025

0.3X = 0.010

X = 0.033 L


In other words, the volume of 0.4 M Ba(OH)2 required is 33 mL.

User LottaLava
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