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s344n2d4d5 [400]
3 years ago
13

In the laboratory a student combines 47.8 mL of a 0.321 M aluminum nitrate solution with 21.8 mL of a 0.366 M aluminum iodide so

lution.
What is the final concentration of aluminum cation ?
M
Chemistry
1 answer:
alex41 [277]3 years ago
4 0

Answer: The final concentration of aluminum cation is 0.335 M.

Explanation:

Given: V_{1} = 47.8 mL (1 mL = 0.001 L) = 0.0478 L

M_{1} = 0.321 M,       V_{2} = 21.8 mL = 0.0218 L,      M_{2} = 0.366 M

As concentration of a substance is the moles of solute divided by volume of solution.

Hence, concentration of aluminum cation is calculated as follows.

[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}

Substitute the values into above formula as follows.

[Al^{3+}] = \frac{M_{1}V_{1} + M_{2}V_{2}}{V_{1} + V_{2}}\\= \frac{0.321 M \times 0.0478 L + 0.366 M \times 0.0218 L}{0.0478 L + 0.0218 L}\\= \frac{0.0153438 + 0.0079788}{0.0696}\\= 0.335 M

Thus, we can conclude that the final concentration of aluminum cation is 0.335 M.

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When NaOH is added the Net ionic reaction would be

  1.    H_3C_6H_5O_7 + 0H^- ----> H_2C_6H_5O_7^- +H_2O \ \ \ \ \ \ \ Main \ Reaction
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  3.    HC_6H_5O_7 ^{2-} + OH^- ---> C_6H_5O_7^{3-} +H_2O \ \ \ \ \ Add \ NaOH  

             

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