What is the pH of a solution made up of 50cm^3 of 1M Ethanoic Acid and 25cm^3 of 1M sodium ethanoate? Assume Ka for Ethanoic acid is 1.8 x 10^-5 mol dm^-3
I'm not quite sure how to do this. I was going to use the equation pH = pKa + log[A-]/[HA] but it doesn't seem to work
You're right. The Henderson-Hasselbalch equation is the one to use. If you want to post your work someone will take a look and try to find the error. You should be substituting 25/75 for for the base and 50/75 for the acid. The denominators cancel, of course, and you have just 25 for (A^-) and 50 for (HA)
To calculate the pH of the solution, you need to consider the acid-base equilibrium between ethanoic acid (HA) and its conjugate base, acetate ion (A-). The equation pH = pKa + log([A-]/[HA]) is indeed the correct equation to use here.
In this case, you have a solution made up of two components: 50 cm3 of 1M ethanoic acid (HA) and 25 cm3 of 1M sodium ethanoate (A-). Since sodium ethanoate is fully dissociated in water, you can assume that you have 25 mmol of acetate ion (A-) in your solution.
First, calculate the number of moles of ethanoic acid (HA) in the 50 cm3 solution:
Number of moles = concentration x volume
Number of moles of HA = 1 M x 50 cm3 = 0.05 moles
Next, calculate the ratio of [A-]/[HA]:
[A-]/[HA] = 25 mmol/0.05 moles = 500
Now, you can substitute the values into the equation pH = pKa + log([A-]/[HA]):
pH = -log(1.8 x 10^-5) + log(500)
To simplify this calculation, you can convert the logarithm with base 10 into natural logarithm (ln) using the formula: log(a) = ln(a)/ln(10).
pH = -log(1.8 x 10^-5) + log(500)
= -(ln(1.8 x 10^-5)/ln(10)) + ln(500)/ln(10)
Now, let's calculate the pH value using a calculator or computer program:
pH ≈ 4.74
So, the pH of the given solution is approximately 4.74.