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3 years ago

What is the approximate pH of a .06 M solution of CH3COOH given that Ka= 1.78*10-5

1 answer:

4 0

CH₃COOH ⇔ CH₃COO⁻ + H⁺

[CH₃COO⁻] = [H⁺] = x

$K=\frac{|CH_{3}COO^{-}|*|H^{+}|}{|CH_{3}COOH|}=\frac{x^{2}}{|CH_{3}COOH|}\\\\
1,78*10^{-5}=\frac{x^{2}}{0,06} \ \ |*0,06\\\\
0,1068*10^{-5}=x^{2}\\\\
x_{1}\approx0,001 \ \land \ \ x_{2}=\approx-0,001\\\\
pH=-log|H^{+}|=-log0,001=3$

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Verdich [7]

Answer:

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Explanation:

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3 years ago

1.24 grams of magnesium phosphate tribasic dissolved in 1 L of lemon juice. What is the Ksp of the magnesium phosphate tribasic

vivado [14]

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Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

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$\text{Molarity of solution}=\frac{1.24g}{262.85g/mol\times 1}\\\\\text{Molarity of solution}=4.72\times 10^{-3}M$

The equation for the ionization of the magnesium phosphate is given as:

$Mg_3(PO_4)_2\leftrightharpoons 3Mg^{2+}+2PO_4^{3-}$

Expression for the solubility product of $Mg_3(PO_4)_2$ will be:

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

We are given:

$[Mg^{2+}]=(3\times 4.72\times 10^{-3})=1.416\times 10^{-2}M$

$[PO_4^{3-}]=(2\times 4.72\times 10^{-3})=9.44\times 10^{-3}M$

Putting values in above expression, we get:

$K_{sp}=(1.416\times 10^{-2})^3\times (9.44\times 10^{-3})^2=2.52\times 10^{-10}$

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3 years ago

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Answer:

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Explanation:

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3 years ago

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Andreas93 [3]

I think you mean:
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3 years ago