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eduard
3 years ago
10

In the experiment "Beer-Lambert’s Law and Spectrophotometry", you prepared a calibration plot similar to the one pictured below.

What is the approximate concentration of a solution whose absorbance is 0.35?
Chemistry
1 answer:
dexar [7]3 years ago
8 0

Answer:

0.025M

Explanation:

As you must see in your graph, each concentration of the experiment has an absorbance. Following the Beer-Lambert's law that states "The absorbance of a solution is directely proportional to its concentration".

At 0.35 of absorbance, the plot has a concentration of:

<h3>0.025M</h3>
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If the ka of a monoprotic weak acid is 5.4 × 10-6, what is the ph of a 0.14 m solution of this acid?
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pH = 2.1

Let HA (aq) resembles the acid; as a weak acid (a small value of K_{a}) HA would partially dissociate to produce protons H^{+} and A^{-}, its conjugate base. Let the final proton concentration (i.e., [H^{+}]) be x. (Apparently x \ge 0) Construct the following RICE table:

\left\begin{array}{cccccc}\text{R}&HA(aq)&\rightleftharpoons&H^{+}(aq) &+ &A^{-}(aq)\\\text{I} & 0.14 \; \text{M} & &\\\text{C}&-x \; \text{M}& &+x \; \text{M} & & +x \; \text{M}\\E & (0.14 - x)\; \text{M} & & x \; \text{M} & &\+x \; \text{M}\end{array}\right

By definition, (all concentrations are under equilibrium condition)

\left\begin{array}{ccc}K_{a}&=&[H^{+}] \cdot [A^{-}] / [HA]\\&=&x^{2} /(0.14 - x)\end{array}\right

It is given that

K_a = 5.4 \cdot 10^{-6}

Equating and simplifying the two expressions gives a quadratic equation; solve the equation for x gives:

x^2 = 5.4 \cdot 10^{-6} \cdot (14 - x) \\x^2 + 5.4 \cdot 10^{-6} \cdot 14 \cdot x - 5.4 \cdot 10^{-6} \cdot 14 = 0 \\x = 0.0087 \; \text{M} \; (x \ge 0)

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\text{pH} = -\text{ln(}[H^{+}]\text{)} / \text{ln(}10\text{)} = 2.1

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