Question

What is the molarity of sodium ions in a solution prepared by mixing 236 ml of 0.75 M sodium phosphate with 252.8 ml of 1.2 M sodium sulfide. Enter to 2 decimal places.

Answer 1

Answer:

2.33 M is the molarity of sodium ions in a solution prepared by mixing.

Explanation:

$Molarity=\frac{n}{V(L)}$

n = moles of substance

V = Volume of solution in L

In, 236 ml of 0.75 M sodium phosphate :

Moles of sodium phosphate = n

Volume of sodium phosphate solution = V = 236 mL = 0.236 L(1 m L =0.001 L)

Molarity of the solution = M = 0.75 M

$n=M\times V=0.75 M\times 0.236 L=0.177 mol$

Sodium phosphate = $Na_3PO_4$

1 mole of sodium phosphate has 3 mol of sodium ions.

Then 0.177 moles will have = 0.177 mol × 3 = 0.531 mol

In, 252.8 ml of 1.2 M sodium sulfide:

Moles of sodium sulfide = n'

Volume of sodium sulfide solution = V' = 252.8 mL = 0.2528 L(1 m L =0.001 L)

Molarity of the solution = M' = 1.2 M

$n'=M'\times V'=1.2 M\times 0.2528 L=0.30336 mol$

Sodium sulfide= $Na_2S$

1 mole of sodium sulfide has 2 mol of sodium ions.

Then 0.30336 moles will have = 0.30336 mol × 2 = 0.60672 mol

After mixing both solutions:

Moles of sodium ions = 0.60672 mol + 0.531 mol = 1.13772 mol

Volume of the mixture = 0.2528 L = 0.236 L = 0.4888 L

Molarity of sodium ions:

$=\frac{1.13772 mol}{0.4888 L}=2.3275 M\approx 2.33 M$

2.33 M is the molarity of sodium ions in a solution prepared by mixing.

Answer 2

The molarity of sodium ion, Na⁺ in the resulting solution is 2.33 M

We'll begin by calculating the number of mole of sodium ion, Na⁺ in each solution.

For Na₃PO₄:

Volume = 236 mL = 236 / 1000 = 0.236 L

Molarity = 0.75 M

Mole of Na₃PO₄ =?

Mole = Molarity x Volume

Mole of Na₃PO₄ = 0.75 × 0.236

Mole of Na₃PO₄ = 0.177 mole

Na₃PO₄(aq) —> 3Na⁺(aq) + PO₄³¯(aq)

From the balanced equation above,

1 mole of Na₃PO₄ contains 3 mole of Na⁺

Therefore,

0.177 mole of Na₃PO₄ will also contain = 0.177 × 3 = 0.531 mole of Na⁺

Thus, 0.531 mole of Na⁺ is present in 480 mL of 0.75 M Na₃PO₄

For Na₂S:

Volume = 252.8 mL = 252.8 / 1000 = 0.2528 L

Molarity = 1.2 M

Mole of Na₂S =?

Mole = Molarity x Volume

Mole of Na₂S = 1.2 × 0.2528

Mole of Na₂S = 0.30336 mole

Na₂S(aq) —> 2Na⁺(aq) + S²¯(aq)

From the balanced equation above,

1 mole of Na₂S contains 2 moles of Na⁺

Therefore,

0.30336 mole of Na₂S will contain = 0.30336 × 2 = 0.60672 mole of Na⁺

Thus, 0.60672 mole of Na⁺ is present in 252.8 mL of 1.2 M Na₂S

• Next, we shall determine the total mole of Na⁺ in the resulting solution.

Mole of Na⁺ in Na₃PO₄ = 0.531 mole

Mole of Na⁺ in Na₂S = 0.60672 mole

Total mole = 0.531 + 0.60672

Total mole = 1.13772 mole

• Next, we shall determine the total volume of the resulting solution

Volume of Na₃PO₄ = 0.236 L

Volume of Na₂S = 0.2528 L

Total volume = 0.236 + 0.2528

Total volume = 0.4888 L

• Finally, we shall determine the molarity of Na⁺ in the resulting solution

Total mole = 1.13772 mole

Total volume = 0.4888 L

Molarity of Na⁺ =?

Molarity = mole / Volume

Molarity of Na⁺ = 1.13772 / 0.4888

Molarity of Na⁺ = 2.33 M

Therefore, the molarity of sodium ion, Na⁺ in the resulting solution is 2.33 M

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