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In: Chemistry

Experiments show that the solubility of alum in 23 mL of 1.2 M KOH plus 10...

Experiments show that the solubility of alum in 23 mL of 1.2 M KOH plus 10 mL of 9 M H2SO4 is about 1.0 g at 1.0 °C and 1.7 g at 6.0 °C. Using your measured solution temperature, estimate the amount of alum left in your chilled solution.

Expert Solution

Hi, the question seems incomplete in a sense that the temperature at which the solubility is required is not given. I suppose, by chilled solution you mean solution at 0 degrees. However, I will try to give answer to the best of my understanding. If you need more information you can always post comment and I will be more than happy to answer to your comments. Judging by the data, you could assume that the solubility curve should follow a linear trend, since you do not have enough data points for a curve. If we plot solubility on y-axis and temperature on x-axis. we have the following data.

since the total volume is 33 mL (23 mL + 10 mL). we can calculate solubility per mL as:

you could calculate the slope of the line as:

, the slope is

if you use the equation for straight line, y = mx + c.

you can put the value of x1 and y1 in this equation to calculate the value of the intercept c. I calculate c as 0.0261.

the line that I drew is

so, you have the line equation y = 0.0042x + 0.0261. if you want to calculate the solubility at 0 degrees, put x=0 and y = 0.0261 g/mL and since you have 33 mL of solution the amount of alum in it is

you can use this equation to calculate the solubility at any temperature between 1 degrees and 6 degrees. I hope it helps. If you need anything else, please post a comment.

queen_honey_blossom answered 2 years ago

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