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almond37 [142]
3 years ago
7

At 100.0∘C, the ion product of water is 5.13×10−13. What is the concentration of hydroxide ions at this temperature? Your answer

should include three significant figures. Write your answer in scientific notation. Use the multiplication symbol rather than the letter x in your answer.
Chemistry
1 answer:
Trava [24]3 years ago
6 0

Answer: The concentration of hydroxide ions at this temperature is 7.16\times 10^{-7}

Explanation:

When an expression is formed by taking the product of concentration of ions raised to the power of their stoichiometric coefficients in the solution of a salt is known as ionic product.

The ionic product for water is written as:

K_w=[H^+][OH^-]

5.13\times 10^{-13}=[H^+][OH^-]

As [H^+]=[OH^-]

5.13\times 10^{-13}=x^2

x=7.16\times 10^{-7}

The concentration of hydroxide ions at this temperature is 7.16\times 10^{-7}

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If 16.9 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc
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Answer: 69.72 kg of cryolite will be produced.

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To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

moles of Al_2O_3 = \frac{16.9\times 1000g}{102g/mol}=165.7moles

moles of NaOH = \frac{57.4\times 1000g}{40g/mol}=1435moles

moles of HF = \frac{57.4\times 1000g}{20g/mol}=2870moles

As 1 mole of Al_2O_3 reacts with 6 moles of NaOH

166 moles of  Al_2O_3 reacts with = \frac{6}{1}\times 166=996 moles of NaOH

As 1 mole of Al_2O_3 reacts with 12 moles of HF

166 moles of  Al_2O_3 reacts with = \frac{12}{1}\times 166=1992 moles of HF

Thus Al_2O_3 is the limiting reagent.

As 1 mole of Al_2O_3 produces = 2 moles of cryolite

166 moles of  Al_2O_3 reacts with = \frac{2}{1}\times 166=332 moles of cryolite

Mass of cryolite (Na_3AlF_6) = moles\times {\text {molar mass}}=332mol\times 210g/mol=69720g=69.72kg

Thus 69.72 kg of cryolite will be produced.

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