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velikii [3]
2 years ago
12

Which one of the following statements is not true concerning 2.00 L of a 0.100 M solution of Ca3(PO4)2?

Chemistry
1 answer:
Len [333]2 years ago
6 0

<u>Answer:</u> The correct answer is Option B.

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .......(1)

  • <u>For A:</u>

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 2.00 L

Putting values in equation 1, we get:

0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{2.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 2.00L)=0.200mol

Moles of calcium phosphate = 0.200 moles

  • <u>For B:</u>

1 mole of calcium phosphate contains 3 moles of calcium atoms, 2 moles of phosphate atoms and 8 moles of oxygen atoms.

So, 0.200 moles of calcium phosphate will contain = (8\times 0.200)=1.6 moles of oxygen atoms.

Moles of oxygen atoms = 1.6 moles

  • <u>For C:</u>

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 1.00 L

Putting values in equation 1, we get:

0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{1.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 1.00L)=0.100mol

Moles of calcium ions = (0.100\times 3)=0.300 moles

  • <u>For D:</u>

Molarity of calcium phosphate solution = 0.100 M

Volume of solution = 5.00 L

Putting values in equation 1, we get:

0.100M=\frac{\text{Moles of }Ca_3(PO_4)_2}{5.00}\\\\\text{Moles of }Ca_3(PO_4)_2=(0.100mol/L\times 5.00L)=0.500mol

Moles of phosphorus atoms = (0.500\times 2)=1.00 moles

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of atoms

Number of phosphorus atoms in 0.500 moles of calcium phosphate will be = (1.00\times 6.022\times 10^{23})=6.022\times 10^{23}

  • <u>For E:</u>

1 mole of calcium phosphate contains 3 moles of calcium ions and 2 moles of phosphate ions.

So, 0.200 moles of calcium phosphate will contain = (3\times 0.200)=0.600 moles of calcium ions

Moles of calcium ions = 0.600 moles

Hence, the correct answer is Option B.

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8 0
2 years ago
What is the pH of a solution that is 0.40 M NaBrO and 0.50 M HBrO (hypobromous acid) (Ka for HBrO = 2.3 x 10^-9)
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Answer

pH=8.5414

Procedure

The Henderson–Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Kₐ. In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution.

pH = pKa + log₁₀ ([A⁻] / [HA])

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[A⁻]= concentration of conjugate base

First, calculate the pKa

pKa=-log₁₀(Ka)= 8.6383

Then use the equation to get the pH (in this case the acid is HBrO)

pH=8.6383+\log_{10}(\frac{0.40\text{ M}}{0.50\text{ M}})=8.5414

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