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3 years ago

What is the pH of a solution with [H+] = 1.25 mc015-1.jpg 10–10 M? Use mc015-2.jpg. –10.1 –9.90 7.90 9.90

Chemistry

2 answers:

kipiarov [429]3 years ago

6 0

We use the relationship:
pH = -log[H⁺]
pH = -log(1.25 x 10⁻¹⁰)

pH = 9.9

Mademuasel [1]3 years ago

4 0

Answer : Option D) 9.90

Explanation : The pH of the solution can be determined by the formula as

pH = - log $[H^{+}]$

Here, the concentration of $[H^{+}]$ = $1.25 X 10^{-10} M$

So, substituting this value in above formula;

we get pH = -log $[ 1.25 X 10^{-10} M]$

Therefore, pH = 9.90.

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2 years ago

A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so

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Answer : The equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

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First we have to calculate the initial moles of $Fe^{3+}$ and $SCN^-$ .

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$\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol$

and,

$\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}$

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$Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)$

Since 1 mole of $Fe^{3+}$ reacts with 1 mole of $SCN^-$ to give 1 mole of $FeSCN^{2+}$

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Now we have to calculate the concentration of $FeSCN^{2+}$ .

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Using Beer-Lambert's law :

$A=\epsilon \times C\times l$

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C = concentration of solution

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$\epsilon$ and l are same for stock solution and dilute solution. So,

$\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}$

For trial solution:

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$[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)$

$[SCN^-]_{eqm}=4.58\times 10^{-8}M$

Therefore, the equilibrium concentration of $SCN^-$ in the trial solution is $4.58\times 10^{-8}M$

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