1.00g of potassium ethanoate is dissolved in 50 cm^3 of 0.200 mol dm^-3 ethanoic acid (Ka = 1.74 * 10^-5 mol dm^-3)
Find the pH.
When you work it out could you please show me step by step how you did it, I know how to use Ka calculations but this one gave me an answer of -1.232 for the pH. ????????????????
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- Thread Starter
- 27-10-2017 19:29
- 27-10-2017 19:33
1) Find the number of mol or potassium ethanoate
2) use the equation, pH = pKa + lg (concentration of salt/ concentration of acid)
- in this case, concentration of salt is the no. Of mol per dm3 of potassium ethanoate
- acid ... just find the no. Of mol (mol = concentration x volume)
- 27-10-2017 19:38