If #15.0mL# of glacial acetic acid (pure #HC_2H_3O_2#) is diluted to #1.50L# with water, what is the pH of the resulting solution? The density of glacial acetic acid is #"1.05 g/mL"#
For acetic acid, the acid dissociation constant is given as
The idea is to dilute a sample of pure acetic acid by a dilution factor of 100. You can calculate the number of grams of acid you're diluting by using the density of the pure acetic acid.
Determine how many moles you would get in that amount of mass by using the acid's molar mass.
This indicates that the molarity of the acetic acid solution is
Next, determine the precise amount of hydronium ions that would result from the acid's dissociation using an ICE Table.
Consequently, the hydronium ion concentration will be
The solution's pH will be the same as
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To find the pH of the resulting solution, first calculate the molarity of acetic acid (HC2H3O2) after dilution. Then, use the dissociation constant of acetic acid to find the concentration of hydronium ions, from which you can determine the pH.

Calculate the number of moles of glacial acetic acid: ( \text{Volume} \times \text{Density} = \text{mass} ) ( 15.0 , \text{mL} \times 1.05 , \text{g/mL} = 15.75 , \text{g} )

Convert grams to moles: ( \text{moles} = \frac{\text{mass}}{\text{molar mass}} ) ( \text{molar mass of HC2H3O2} = 60.05 , \text{g/mol} ) ( \text{moles} = \frac{15.75 , \text{g}}{60.05 , \text{g/mol}} = 0.262 , \text{mol} )

Calculate the molarity of acetic acid after dilution: ( \text{Molarity} = \frac{\text{moles}}{\text{volume in liters}} ) ( \text{Molarity} = \frac{0.262 , \text{mol}}{1.50 , \text{L}} = 0.175 , \text{M} )

Calculate the concentration of hydronium ions using the dissociation constant of acetic acid (( K_a = 1.8 \times 10^{5} )): ( K_a = \frac{[\text{H}^+][\text{C}_2\text{H}_3\text{O}_2^]}{[\text{HC}_2\text{H}_3\text{O}_2]} ) ( \text{Let } x = [\text{H}^+] ) ( \text{Let } x = [\text{C}_2\text{H}_3\text{O}_2^] ) (since acetic acid is a weak acid, the amount dissociated is negligible compared to the initial concentration) ( 1.8 \times 10^{5} = \frac{x \times x}{0.175} ) Solve for ( x ), which represents the concentration of hydronium ions.

Calculate pH using the concentration of hydronium ions: ( \text{pH} = \log([\text{H}^+]) )
Plug in the value of ( x ) to find the pH of the resulting solution.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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