What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water?
This is in regards to my previous question. We are still calculating the pH of w/HCl and w/HC2H3O2.
I don't really understand what to do with the 100ml.
What they want you to do is to take the solution after reaction, dilute to 100 mL, and recalculate the pH.
For the first one, you had 0.00061 moles HCl left over and dilute to 100 mL means that the molarity, instead of being 0.00061/0.0099 = 0.0616 M, it now is 0.00061/0.1 = 0.0061 and pH = 2.21 (as opposed to 1.21 earlier).
For the second one,
you had 0.00019 of the salt, and 0.00061 moles of the acetic acid left. Instead of 0.00019/0.0099 and 0.00061/0.0099 the concns will be 0.00019/0.1 = ?? and 0.00061/0.1 = ?? The final pH, if I didn't goof will be 4.23 (as opposed to 4.23 earlier; i.e., no change).
You may wonder what all of this is about so don't miss the point of these questions. When you took the first one (strong acid/strong base) and diluted the excess acid the pH changed BIG TIME from 1.21 to 2.21 (a change of 10 in acidity) BUT doing the same thing to the weak acid/weak base (the buffered solution) THERE IS NO CHANGE IN pH even though you have added 90 mL MORE water to it. Remember that a buffered solution is supposed to resist a change in pH AND IT DID.
thank you
To calculate the pH of the diluted acid solution, you need to consider the dilution factor introduced by adding 100 mL of water to the 8.00 mL of 0.10 M acid.
First, you can calculate the moles of acid initially present in the solution using the formula:
moles = concentration (M) x volume (L)
Given that the initial volume of acid is 8.00 mL and the concentration is 0.10 M, you can convert the volume to liters by dividing it by 1000:
moles = 0.10 M x (8.00 mL / 1000 mL/L) = 0.008 mol
The dilution factor represents the ratio of the final volume to the initial volume. In this case, the final volume is the sum of the initial volume of acid (8.00 mL) and the added volume of water (100 mL), which equals 108 mL or 0.108 L.
The dilution factor, D, can be calculated as:
D = (final volume) / (initial volume) = 0.108 L / 0.008 L = 13.5
Since dilution is a proportional process, the concentration of the acid after dilution is equal to the initial concentration divided by the dilution factor:
final concentration = initial concentration / dilution factor = 0.10 M / 13.5 = 0.00741 M
Now that you have the final concentration of the acid after dilution, you can use this to calculate the pH.
For a strong acid like HCl, which dissociates completely in water, the concentration of H+ ions is equal to the concentration of the acid. Therefore, pH is calculated as the negative logarithm (base 10) of the H+ ion concentration.
pH = -log10([H+])
So, for HCl with a final concentration of 0.00741 M:
pH = -log10(0.00741)
Using a calculator, you can find that the pH of this solution is approximately 2.13.
For HC2H3O2, which is a weak acid that partially ionizes in water, you need to consider the equation for the dissociation of the acid:
HC2H3O2 ⇌ H+ + C2H3O2-
The equilibrium expression for this dissociation can be written as:
Ka = [H+][C2H3O2-]/[HC2H3O2]
Given that the final concentration of HC2H3O2 is 0.00741 M and that it only partially ionizes, let's assume x mol/L of HC2H3O2 dissociates. The concentration of H+ ions and C2H3O2- ions will be x mol/L, and the concentration of undissociated HC2H3O2 will be (0.00741 - x) mol/L.
Using the equilibrium expression, you can set up the equation:
Ka = (x)(x)/(0.00741 - x)
Given that Ka for HC2H3O2 is 1.7 x 10^-5, you can solve this equation to find the value of x, which will correspond to the concentration of H+ ions. Once you have the concentration of H+, you can use the same equation as before to calculate the pH.