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)
Answer:
for the process is -375 kJ
Explanation:
- Given reaction is a combination of the two given elementary steps.
- Summation of change in standard enthalpy ()of the two elementary reactions give of the reaction .
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<span>B. The ignition point for a given fuel is met.</span>
Your answer would be D, single replacement
Answer: Option (3) is the correct answer.
Explanation:
In an aqueous solution, a compound exists in the form of ions that is in the form of protons and electrons.
Protons are positively charged and electrons are negatively charged. A proton (usually hydrogen ion, ) in water (solvent) also exists as .
For example,
This shows that the conjugate acid of is .
Thus, we can conclude that a hydrogen ion, H+, in aqueous solution may also be written as .