Respuesta :
Answer:
2.33 M is the molarity of sodium ions in a solution prepared by mixing.
Explanation:
[tex]Molarity=\frac{n}{V(L)}[/tex]
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
[tex]n=M\times V=0.75 M\times 0.236 L=0.177 mol[/tex]
Sodium phosphate = [tex]Na_3PO_4[/tex]
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
[tex]n'=M'\times V'=1.2 M\times 0.2528 L=0.30336 mol[/tex]
Sodium sulfide= [tex]Na_2S[/tex]
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:
[tex]=\frac{1.13772 mol}{0.4888 L}=2.3275 M\approx 2.33 M[/tex]
2.33 M is the molarity of sodium ions in a solution prepared by mixing.
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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