First, we have to get moles of CH3COONa = mass/molar mass
= 20 g / 82.03 g/mol = 0.244 moles
when we have [CH3COOH] = 0.15 M
∴ [CH3COONa] = moles of CH3COONa / Volume of solution
= 0.244 moles / 0.5L = 0.488 M
when we look up for Ka of acetic acid value it is equal 1.8 x 10^-5
So we can get Pka = -㏒Ka
= -㏒(1.8 x10^-5)
= 4.7
now we will use Henderson - Hasselbalchn equation to get the PH:
PH = Pka + ㏒[conjugate basic/weak acid]
when CH3COOH is the weak acid & CH3COO- is the conjugate base so by substitution:
PH = 4.7 + ㏒ (0.488/0.15)
= 5.2
b) when we have this equation for the reaction:
HCl + CH3COONa → CH3COOH + NaCl
ionic equation : H+ + Cl- + CH3COO- + Na+ → CH3COOH + Na+ + Cl-
when HCl + H2O → H3O+ + Cl-
∴ the reaction will be:
CH3COO- (aq) + H3O+(aq) → CH3COOH(aq) + H2O(l)
The pH of the buffer is 6.1236.
Explanation:
The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.
The pH of the buffer can be known as
The concentration of
Similarly, the concentration of [HA] =
Then the pH of the buffer will be
pH = 6.247 + log [ 0.304/0.404]
So, the pH of the buffer is 6.1236.
Sodium chloride I.E. NaCl is an ionic compound formed by the reaction between sodium and chlorine. it is also known as common salt .
Evaporation,condensation, and precipitation.