What is the ratio of moles of solute to liters of solution called?

15.0 mL of 0.050 M Ba(NO3)2 M and 100.0 mL of 0.10 M KIO3 are added together in a 250 mL erlenmeyer flask. In this problem, igno

timama [110]

Answer:

Yes, precipitation of barium iodate will occur.

Explanation:

Molarity of barium nitrate solution = 0.050 M

Volume of barium nitrate solution =15.0 mL = 0.0150 L

1 mL = 0.001 L

Moles of barium nitrate = n

$n=0.050 M\times 0.0150L=0.00075 mol$

$Ba(NO_3)_2(aq)\rightarrow Ba^{2+}(aq)+2NO_3^{-}(aq)$

Moles of barium ions: $1\times 0.00075 mol=0.00075 mol$

Molarity of potassium iodate solution = 0.10 M

Volume of potassium iodate solution =100.0 mL = 0.1000 L

1 mL = 0.001 L

Moles of potassium iodate = n'

$n'=0.10 M\times 0.1000 L=0.01 mol$

$KIO_3(aq)\rightarrow K^{+}(aq)+IO_3^{-}(aq)$

Moles of iodate ions = $1\times 0.01 mol=0.01 mol$

After mixing of both solution in 250 mL in erlenmeyer flask

Volume of the final solution = 250 mL = 0.250 L

Concentration of barium ions in 250 mL solution :

$[Ba^{2+}]=\frac{0.00075 mol}{0.250 L}=0.003 M$

Concentration of iodate ions:

$[IO_3^{-}]=\frac{0.01 mol}{0.250 L}=0.04 M$

Solubility product of barium iodate, $K_{sp}=4.01\times 10^{-9}$

Ionic product of the barium iodate in solution : $K_i$

$Ba(IO_3)_2\rightleftahrpoons Ba^{2+}+2IO_3^{-}$

$K_i=[Ba^{2+}][IO_2^{-}]^2$

$K_i=0.003 M\times (0.04 M)^2=4.8\times 10^{-6}$

$K_{sp}$ ( precipitation)

As we can see, the ionic product of the barium iodate is greater than the solubility product of the barium iodate precipitation of barium iodiate will occur in 250 mL of final solution.

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The Dome of the Rock and Al-Aqsa Mosque are located in what city?

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They are located in Jerusalem.

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Read 2 more answers

Which of the following is an example of a polymer?

svetlana [45]

<span>Sucrose </span>is an example of a polymer separate but equal means that both races get the same thing they just dont get it at the same time.<span />

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3 years ago

What is the daughter nucleus of pt-199 after a beta decay?

Setler79 [48]

B. At-200

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3 years ago

H2CO3(aq) + H200 H30 (aq) + HCO3 (aq).

timofeeve [1]

Answer:

$K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}$

Explanation:

Several rules should be followed to write any equilibrium expression properly. In the context of this problem, we're dealing with an aqueous equilibrium:

an equilibrium constant is, first of all, a fraction;
in the numerator of the fraction, we have a product of the concentrations of our products (right-hand side of the equation);
in the denominator of the fraction, we have a product of the concentrations of our reactants (left-hand side o the equation);
each concentration should be raised to the power of the coefficient in the balanced chemical equation;
only aqueous species and gases are included in the equilibrium constant, solids and liquids are omitted.

Following the guidelines, we will omit liquid water and we will include all the other species in the constant. Each coefficient in the balanced equation is '1', so no powers required. Multiply the concentrations of the two products and divide by the concentration of carbonic acid:

$K_a=\frac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}$

4 0

3 years ago