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3 years ago

What is the solubility product (or ion-product), Ksp, expression at equilibrium for calcium phosphate, Ca3(PO4)2?

Chemistry

1 answer:

Gnoma [55]3 years ago

4 0

Answer : The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}]^3[PO_4^{3-}]^2$

Explanation :

Solubility product : It is defined as the product of the concentration of the ions that present in a solution raised to the power by its stoichiometric coefficient in a solution of a salt. This takes place at equilibrium only.

The solubility product constant is represented as, $K__{sp}$ .

The dissociation of calcium phosphate is written as:

$Ca_3(PO_4)_2\rightleftharpoons 3Ca^{2+}+2PO_4^{3-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}]^3[PO_4^{3-}]^2$

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1 year ago

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3 years ago

Read 2 more answers

Acetic acid has a pKa of 4.74. Buffer A: 0.10 M HC2H3O2, 0.10 M NaC2H3O2 Buffer B: 0.30 M HC2H3O2, 0.30 M NaC2H3O2 Buffer C: 0.5

e-lub [12.9K]

Answer:

Buffer B has the highest buffer capacity.

Buffer C has the lowest buffer capacity.

Explanation:

An effective weak acid-conjugate base buffer should have pH equal to $pK_{a}$ of the weak acid. For buffers with the same pH, higher the concentrations of the components in a buffer, higher will the buffer capacity.

Acetic acid is a weak acid and $CH_{3}COO^{-}$ is the conjugate base So, all the given buffers are weak acid-conjugate base buffers. The pH of these buffers are expressed as (Henderson-Hasselbalch):

$pH=pK_{a}(CH_{3}COOH)+log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}$

$pK_{a}(CH_{3}COOH)=4.74$

Buffer A: $pH=4.74+log(\frac{0.10}{0.10})=4.74$

Buffer B: $pH=4.74+log(\frac{0.30}{0.30})=4.74$

Buffer C: $pH=4.74+log(\frac{0.10}{0.50})=4.04$

So, both buffer A and buffer B has same pH value which is also equal to $pK_{a}$ . Buffer B has higher concentrations of the components as compared to buffer A, Hence, buffer B has the highest buffer capacity.

The pH of buffer C is far away from $pK_{a}$ . Therefore, buffer C has the lowest buffer capacity.

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3 years ago

A balloon contains 11.1 g of nitrogen monoxide gas (NO). how many molecules are in the balloon?

grandymaker [24]

The number of molecules that are in balloon are = 2.227 x10^23 molecules

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calculate the number of moles of NO

moles = mass/molar mass

molar mass of NO = 14+ 16 = 30 g/mol

moles is therefore= 11.1 g/30g/mol= 0.37 moles

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0.37 =? molecules

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