Question

A 15.5 mL sample of 0.215 M KOH solution requires 21.2 mL of acetic acid in a titration experiment. Calculate the molarity of the acetic acid solution. Please show all steps, I don't understand this.
CH3COOH + KOH ==> CH3COOK + H2O
moles KOH = M x L = ??
Convert moles KOH to moles CH3COOH. Since the coefficients are one in the equation, moles KOH = moles CH3COOH.
Then M acid = moles acid/L acid.
the concentration of HCl, a strong acid is 0.500 M. if 20.0 ml HCl is needed to titrate 40.0 ml of NaOH, what is the concentration of NaOH
I don't understand how to do this. Can I see a step by step way of how to do this please

Related Questions

You are carrying out the titration of 100.0 mL of 1.000M acetic acid with 1.000 M sodium hydroxide.... A 10.0 ml sample of vinegar which is an aqueous solution of acetic acid, requires 16.5ml of a 0.500M... A 21.7 mL sample of 0.18 M KOH solution required 26.3 mL of aqueous acetic acid solu- tion in a titr...