Answer: 75%
Explanation:
The following information can be gotten from the question:
Waste = 70kg
Theoretical yield = 280kg
Therefore, the actual yield will be the difference between the theoretical yield and the waste which will be:
= 280kg - 70kg = 210kg
The percent yield will now be:
= Actual yield / Theoretical yield × 100
= 210/280 × 100
= 3/4 × 100
= 75%
We are given the molar concentration of an aqueous solution of weak acid and the pH ofthe solution, and we are asked to determine the value of Ka for the acid.
The first step in solving any equilibrium problem is to write the equation for the equilibriumreaction. The ionization of benzoic acid can be written as seen in the attached image (1).
The equilibrium-constant expression is the equation number (2)
From the measured pH, we can calculate pH as seen in equation (3)
To determine the concentrations of the species involved in the equilibrium, we imagine that thesolution is initially 0.10 M in HCOOH molecules. We then consider the ionization of the acidinto H+ and HCOO-. For each HCOOH molecule that ionizes, one H+ ion and one HCOO- ionare produced in solution. Because the pH measurement indicates that [H+] = 1x 10^-4 M atequilibrium, we can construct the following table as seen in the equation number (4)
To find the value of Ka, please see equation (5):
We can now insert the equilibrium concentrations into the expression for Ka as seen in equation (6)
Therefore, 2.58x10^-4 M is the concentration of benzoic acid to have a pH of 4.0
Answer:
4.48
Explanation:
p[OH-] = - log(3.00 *10^-10) = 9.52
pH + pOH = 14
pH = 14 - 9.52= 4.48
The volume of the NaOH used is calculated as 14 mL.
<h3>What is stoichiometry?</h3>
The term stoichiometry has to do with the calculation of the amount of substance in a reaction using mass - mole or mass - volume relationship.
Here;
Number of moles of CaCO3 = 0.205 g/100.1 = 0.00205 moles
Number of moles of HCl = 2.00 M * 7/1000 L = 0.014 moles
2 moles of HCl reacts with 1 mole of CaCO3
x moles of HCl reacts with 0.00205 moles of CaCO3
x = 0.00205 moles * 2/1 = 0.0041 moles
Hence HCl is the excess reactant
Amount of excess HCl = 0.014 moles - 0.0041 moles = 0.0099 moles
Concentration of excess HCl reacted = 0.0099 moles/125 * 10^-3 = 0.0792 M
Using;
CAVA/CBVB = NA/NB
CAVANB = CBVBNA
VB = CAVANB/CBNA
VB = 0.0792 M * 10 mL * 1/ 0.058 M
VB = 14 mL
Missing parts;
A 0.205 g sample of caco3 (mr = 100.1 g/mol) is added to a flask along with 7.50 ml of 2.00 m hcl. caco3(aq) + 2hcl(aq) → cacl2(aq) + h2o(l) + co2(g) enough water is then added to make a 125.0 ml solution. a 10.00 ml aliquot of this solution is taken and titrated with 0.058 m naoh. naoh(aq) + hcl(aq) → h2o(l) + nacl(aq) how many ml of naoh are used?
Learn more about stoichiometry: brainly.com/question/9743981