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3 years ago

NH4NO3, (aq) + Ba(OH)2(aq) →

Chemistry

1 answer:

mestny [16]3 years ago

3 0

Answer:

Ba(OH)2 (aq) + 2 NH4NO3 (aq) → 2 NH4OH (l) + Ba(NO3)2 (aq)

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The Ksp of barium fluoride, BaF2, is 1.6 x 10-6. A solution of 5.0M NaF is added dropwise to a 2.0L solution that is 0.016 M in

Hatshy [7]

Answer:

Precipitation will occur after the concentration of fluoride ions exceeds 0.01 M.

4 mL volume of NaF must be added to cause $BaF_2$ to precipitate.

Explanation:

Concentration of barium ions = $[Ba^{2+}]=0.016 M$

Volume of barium ion solution = 2.0 L

$BaF_2\rightleftharpoons Ba^{2+}+2F^-$

The solubility product of the barium fluoride = $K_{sp}=1.6\times 10^{-6}$

$K_{sp}=[Ba^{2+}][F^-]^2$

$1.6\times 10^{-6}=[0.016 M]\times [F^-]^2$

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Precipitation will occur after the concentration of fluoride ions exceeds 0.01 M.

Concentration of fluoride ion = $C_1=0.01 M$

Volume of solution = $V_1=2.0 L$

Concentration of NaF solution added = $C_2=5.0 M$

Volume of NaF solution added = $V_2=?$

$C_1V_1=C_2V_2$

$V_2=\frac{0.01 M\times 2.0 L}{5.0 M}=0.004L$

1 L = 1000 mL

0.004 L = 0.004 × 1000 mL = 4 mL

4 mL volume of NaF must be added to cause $BaF_2$ to precipitate.

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Answer:

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