NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.
Answer:
Approximately
.
Explanation:
Make use of the molar mass data (
) to calculate the number of moles of molecules in that
of
:
.
Make sure that the equation for this reaction is balanced.
Coefficient of
in this equation:
.
Coefficient of
in this equation:
.
In other words, for every two moles of
that this reaction consumes, two moles of
would be produced.
Equivalently, for every mole of
that this reaction consumes, one mole of
would be produced.
Hence the ratio:
.
Apply this ratio to find the number of moles of
that this reaction would have produced:
.
Hmm... I'm unsure but its either B or D.
Kc = [H3O+][HCO3-] / [H2CO3]
Remember that Kc is products over reactants. Also, you do not include liquid water in a Kc expression, since liquid water has no concentration.
Answer:
C
Explanation:
C because there is more in it than the rest of the idems