pH = -log10(Ka) - log10([HA]/[A-])= 4.20 - log10([HA]/[A-])
in which
and
So pH = 4.20 - log10([2.5]/[1.8]) = 4.20 - log10(1.39) = 4.20 - 0.14 = 4.06
There can be no such reaction. Perhaps you meant HCl(aq) + C6H5COONa to get C6H5COOH plus NaCl. That would be converting the sodium salt of benzoic acid (sodium benzoate) into benzoic acid and sodium chloride by using hydrochloric acid.
only the benzoic acid C6H5COOH
You're starting with something like sodium benzoate (depends on what base you used for the extraction step), which contains sodium ions (Na+) and benzoate ions (C6H5COO-). Ionic compounds like that tend to be soluble in water. When you add H+, you protonate the benzoate ion to make benzoic acid (C6H5COOH), which is a neutral molecule, and hardly soluble in water at all... so it precipitates.
BENZOATE
Benzoic acid has a chemical formula of PhCOOH. It does not react with water so much as it dissociates in it, forming a free hydrogen ion. The reaction for the dissociation is PhCOOH(aq) --> PhCOO-(aq) + H+ (aq).
C6H5COOH + NaHCO3 ===> C6H5COO-Na+ + H2O + CO2
The reaction for benzoic acid and methyl amine produces benzamide. The equation is C6H5COOH + CH3NH2 ---> C6H5CONHCH3 + H2O.
There can be no such reaction. Perhaps you meant HCl(aq) + C6H5COONa to get C6H5COOH plus NaCl. That would be converting the sodium salt of benzoic acid (sodium benzoate) into benzoic acid and sodium chloride by using hydrochloric acid.
Sodium Benzoic Acid
only the benzoic acid C6H5COOH
You're starting with something like sodium benzoate (depends on what base you used for the extraction step), which contains sodium ions (Na+) and benzoate ions (C6H5COO-). Ionic compounds like that tend to be soluble in water. When you add H+, you protonate the benzoate ion to make benzoic acid (C6H5COOH), which is a neutral molecule, and hardly soluble in water at all... so it precipitates.
Benzoic acid has a chemical formula of PhCOOH. It does not react with water so much as it dissociates in it, forming a free hydrogen ion. The reaction for the dissociation is PhCOOH(aq) --> PhCOO-(aq) + H+ (aq).
BENZOATE
The reaction is the following:Mg(OH)2 + 2 C6H5COOH = Mg(C6H5COO)2 + 2 H2O
C6H5COOH + NaOH + I2 -----------> C6H5COOI + NaI + H2O
K2CO3 + 2 benzoic acid -> 2 potassium benzoate + H2CO3
benzoic acid + Sodium Hydroxide ==> water + sodium benzoate