`CH_3COOH+NaOH rarr CH_3COONa+H_2O`
Let us say we have 1L of `CH_3COOH ` solution.
According to the stoichiometry;
`CH_3COOH:NaOH = 1:1`
`CH_3COOH:CH_3COONa = 1:1`
Since we are adding enough NaOH full reaction will be occurred. We can assume that there is no partial reactions.
CH_3COOH moles present before tritration `= 0.018`
NaOH moles required for titration `= 0.018`
Required NaOH volume from 0.036M `= 1/0.036*0.018 = 0.5`
`CH_3COOH:CH_3COONa = 1:1`
Final CH_3COONa mole in the solution `= 0.018`
Final volume of the solution` = 1+0.5 = 1.5L`
`CH_3COONa rarr CH_3COO^-+Na^+`
`[CH_3COO^-] = 0.018/1.5 = 0.012M`
So the concentration of `CH_3COO^-` ions is 0.012M.
We’ll help your grades soar
Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.
- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support
Already a member? Log in here.
Are you a teacher? Sign up now