Answer:
53.9 g
Explanation:
We must use the Henderson-Hasselbach equation to answer this question:
pH = pKa + log [A⁻]/[HA]
we know the pH, pKa (pKa = -log Ka), thus we can compute the ratio [A⁻]/[HA], and from this the mass of KC6H5CO2 knowing that M = mol/L.
therefore,
4.63 = - log ( 6.3 x 10⁻⁵ ) + log [A⁻]/[HA]
4.63 = - ( - 4.20 ) + log [A⁻]/[HA]
0.43 = log [A⁻]/[HA]
taking inverse log to both sides of this equation
2.69 = [A⁻]/[HA]
Now [A⁻] =2.69 x [HA] =2.69 x 1.00 M = 2.69 M
We know the molarity is equal to mol per liter of solution, so
mol KC6H5CO2 = 2.69 mol/L x 0.125 L = 0.36 mol
and using the molecular weight of KC6H5CO2 we get that the mass is
0.336 mol x 160.21 g/mol = 53.9 g
The student should take 53.9 g of KC6H5CO2
Answer:
He should dissolve 53.9 grams of KC6H5CO2
Explanation:
Step 1: Data given
Volume of A 1.00 M benzoic acid solution = 125 mL = 0.125 L
Ka of benzoic acid = 6.3*10^-5
pH = 4.63
Step 2: Calculate concentration of conjugate base
pH = pKa + log ([A-]/[HA])
4.63 = 4.20 + log ([A-]/[HA])
0.43 = log ([A-]/[HA])
10^0.43 = [A-]/[HA])
2.69 = [A-]/[HA])
2.69 = [A-]/ 1.00
[A-] = 2.69 M
Step 3: Calculate moles KC6H5CO2
Moles molarity * volume
Moles = 2.69 M * 0.125 L
Moles = 0.33625 moles
Step 4: Calculate mass KC6H5CO2
Mass of KC6H5CO2 = moles * molar mass
Mass of KC6H5CO2 = 0.33625 moles * 160.21 g/mol
Mass of KC6H5CO2 = 53.9 grams
He should dissolve 53.9 grams of KC6H5CO2
