Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
Answer:
B. Molarity will decrease
Explanation:
Molarity is one of the measures of the molar concentration of a solution. It is calculated by dividing the number of moles of the solute by the volume of the solvent. This means that the higher the amount of solute in relation to the volume of solvent, the higher the molarity of that solution.
In essence, adding water to a solution dilutes it i.e it increases the solvent's volume in relation to the solute, causing the molarity to decrease. In a nutshell, diluting a solution (by adding water or more solvent) causes the molarity of such solution to decrease. For example, if water is added to a 0.70 molar solution of CuSO4, the molarity of the solution will DECREASE.
Answer:
See explanation
Explanation:
The reaction equation is;
C3H8 (g) + 5O2(g) -------> 4H2O(g) + 3CO2(g)
From the formula;
Total enthalpy of reactants = (ΔHf of Reactant 1 x Coefficient) + (ΔHf of Reactant 2 x Coefficient)
Total enthalpy of products= (ΔHf of Product 1 x Coefficient) + (ΔHf of Product 2 x Coefficient)
Hence;
Total enthalpy of reactants =[(-103.85 * 1) + (0 * 5)] = -103.85 + 0 = -103.85 KJ/mol
Total enthalpy of products= [(-393.51 * 4) +(-241.82 * 3)] = (-1574.04) + (-483.64) = -2057.68 KJ/mol
Explanation:
A freezing point is defined as the point in which a liquid state of a substance changes into solid state.
Whereas boiling point is defined as the point at which liquid state of substance starts to convert into vapor state.
So, a substance is freezing at 
and boils at 
then it means at room temperature, that is, around 
the substance is present in liquid state.
This is because between the freezing and boiling point a substance will always exist in liquid state.
Thus, we can conclude that state of the material at room temperature is liquid.
