of 0.1 M NaOH with 0.2213 M maleic acid
1. First Ionization: H2C4H4O4 + 2NaOH → 2H2O + Na2C4H4O4
Second Ionization: Na2C4H4O4 + 2NaOH → 2NaOH + Na2C4H4O4
2. The molecular mass of maleic acid is 116.07 g/mol.
h2c4h4o4 + 2naoh --> 2h2o + na2c4h4o4
as for the previous question im sorry i forgot to plug the maleic acid in the formula properly. so this ones it ^
.2213 of c4h4o4 used (maleic acid)
and
.1 M NaOH
1. show the equation for 1st and 2nd ionization
2. find molecular mass of maleic acid
help? How do I do this?
I have a list of V (mL) and pH values...
basically titrated 50 mL
1 answer