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SSSSS [86.1K]
3 years ago
7

You wish to make a 0.299 M hydroiodic acid solution from a stock solution of 6.00 M hydroiodic acid. How much concentrated acid

must you add to obtain a total volume of 50.0 mL of the dilute solution?
Chemistry
1 answer:
Art [367]3 years ago
5 0

Answer:

V_1=2.49mL

Explanation:

Hello,

In this case, considering that the moles of hydrioiodic acid remain unchanged during the dilution process:

n_{HI}=n_{HI}

One could apply the following equality in terms of molarity:

V_1M_1=V_2M_2

Whereas the subscript 1 accounts for the solution before the dilution and 2 after the dilution, therefore, the required volume of 6.00 M acid is:

V_1=\frac{V_2M_2}{M_1} =\frac{50.0mL*0.299M}{6.00M}=2.49mL

Best regards.

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3 years ago
K2so4 is a strong electrolyte. determine the concentration of each of the individual ions in a 0.450 m k2so4 solution.
amid [387]

<em>Answer:</em>

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<em>Explanation:</em>

<em>Data Given:</em>

Conc. of H2SO4 = 0.450

As sulphoric acid is a strong electrolyte, it completely dissociate into ions.

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.450 M K2SO4 means that there is .450 mols of K2SO4 in every liter of solution.

                      K2SO4  :  K+                                          K2SO4  : SO4∧-2

                          1      =     2                                                   1     =    1

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Conc. of potassium ion will be 0.90M    

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