Calculate the molar concentration of H+ ion of a solution that is 3.45×10-1 M in CH3CO2H and 1.62×10-3 M in the salt CH3CO2Li.
CH3CO2H(aq) = CH3CO2-(aq) + H+(aq)
Can't you use the Henderson-Hasselbalch equation?
To calculate the molar concentration of H+ ion in the solution, you need to consider the ionization of CH3CO2H to calculate the concentration of H+ ions produced.
First, let's calculate the concentration of H+ ions from CH3CO2H (acetic acid):
The chemical equation for the dissociation of acetic acid is CH3CO2H(aq) → CH3CO2-(aq) + H+(aq)
Since acetic acid is a weak acid, it does not completely ionize. Thus, only a fraction of CH3CO2H molecules will dissociate into H+ ions and CH3CO2- ions.
The molar concentration of H+ ions from CH3CO2H can be calculated using the equilibrium constant expression, Ka:
Ka = [H+][CH3CO2-]/[CH3CO2H]
Since the concentration of CH3CO2H is given as 3.45×10^-1 M, and the concentration of CH3CO2- can be assumed to be negligible compared to the initial concentration of CH3CO2H, we can simplify the equation:
Ka = [H+][CH3CO2-]/[CH3CO2H] ≈ [H+]/[CH3CO2H]
Since the concentration of H+ is what we want to find, we can manipulate the equation to solve for [H+]:
[H+] = Ka * [CH3CO2H]
Now, we need the Ka value for acetic acid. The Ka value can be found in a reference table or by looking it up in a chemistry database. The Ka value for acetic acid is approximated to be 1.8 × 10^-5.
[H+] = (1.8 × 10^-5) * (3.45 × 10^-1)
Calculating this expression, we get:
[H+] ≈ 6.21 × 10^-6 M
So, the molar concentration of H+ ions from CH3CO2H is approximately 6.21 × 10^-6 M.
Now, let's calculate the molar concentration of H+ ions from the salt CH3CO2Li:
Since CH3CO2Li is a salt, it dissociates completely in water, giving equal concentrations of CH3CO2- and Li+ ions.
The concentration of H+ ions from CH3CO2Li will be negligible, as Li+ ions do not have a significant effect on acidity. Therefore, the concentration of H+ ions from CH3CO2Li is essentially zero.
Finally, to find the total molar concentration of H+ ions in the solution, we need to sum the concentrations from CH3CO2H and CH3CO2Li:
Total [H+] = [H+] from CH3CO2H + [H+] from CH3CO2Li ≈ 6.21 × 10^-6 M + 0 M
Total [H+] ≈ 6.21 × 10^-6 M
Therefore, the molar concentration of H+ ions in the given solution is approximately 6.21 × 10^-6 M.