Valence.
The electrons in the outer shell of an atom are called valence electrons.
Valence electrons determine whether the an element is ready form compounds. These electrons can be gained, lost, or shared in the formation of compounds.
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.
Write an balance the equation
Na2O + H2O -> 2 NaOH
Calculate the molecular mass of Na2O and NaOH from the atomic mass from the periodic table.
Na = 23
O=16
H=1
Na2O = 23 * 2 + 16 = 62
NaOH = 23+16+1= 40
For the stoichiometry of the reaction one mole of Na2O = 62g produce two mol of NaOH = 2* 40= 80 g
120 g Na2O x 80g NaOH / 62g Na2O=
154.8 g NaOH
In comparison see it is very easy in goolge
