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AVprozaik [17]
3 years ago
13

How do I do this whole problem?

Chemistry
1 answer:
KatRina [158]3 years ago
4 0

The solubility of a substance in water is dependent on the temperature. Thus for

1 & 2: Temperature is the independent variable (the one that changes in the first place) and Solubility is a dependent variable (a variable that changes in response to changes in the independent variable.)

The graph: by convention you shall label the horizontal axis with the independent variable and the vertical axis with the dependent variable. For clarity's sake you shall use the finest scale possible that accommodates for all data provided for both axis. Plot the data points on the graph as if they are points on a cartesian plane.

My teacher made no detailed requirements on the phrasing on titles of solubility curve plots; however, like most other graphs in chemistry, the title shall specify the name of variables presented in this visualization. For instance, "the solubility of KClO_3 under different temperatures" might do. You shall refer to your textbooks for such convention.

It is necessary to interpolate to find the solubility at a temperature not given in the table. Start by connecting all given data points with a smooth line; find the vertical line corresponding to temperature = 75 degree Celsius and determine the solubility at the intersection of the vertical line and the trend line. That point shall approximates the solubility of the salt at that temperature.

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Answer:

The pH of the final solution is 7.15

Explanation:

50 mL of 2.0 M of K_2HPO_4 and 25 mL of 2.0 M of KH_2PO_4 were mixed to make a solution

Final volume of the solution after dilution = 200 mL

Final concentration of K_2HPO_4, [K_2HPO_4] = \frac{50 mL\times 2 M}{200 mL} = 0.5 M

Final concentration ofKH_2PO_4, [KH_2PO_4] = \frac{25 mL\times 2 M}{200 mL} = 0.25 M

We use Hasselbach- Henderson equation:

pH = pK_a+ log \frac{[salt]}{[acid]}pka of KH_2PO_4 = 6.85

Substituting the values:

pH = 6.85+ log \frac{0.5}{0.25}pH = 6.85+ log 2pH = 6.85+ 0.3 = 7.15

Therfore,  the pH of the final solution is 7.15

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