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:
2. Inorganic
Explanation:
All man-made and most carbon-based compounds are inorganic
Answer:
0.342 m
Explanation:
From the question given above, the following data were obtained:
Mass of NaBr = 14.57 g
Mass of water = 415 g
Molar mass of NaBr = 102.89 g/mol
Molality of NaBr =?
Next, we shall determine the number of mole in 14.57 g of NaBr. This can be obtained as follow:
Mass of NaBr = 14.57 g
Molar mass of NaBr = 102.89 g/mol
Mole of NaBr =?
Mole = mass / molar mass
Mole of NaBr = 14.57 / 102.89
Mole of NaBr = 0.142 mole
Next, we shall convert 415 g of water to kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
415 g = 415 g × 1 Kg / 1000 g
415 g = 0.415 Kg
Thus, 415 g is equivalent to 0.415 Kg.
Finally, we shall determine Molality of the solution as follow:
Mole of NaBr = 0.142 mole
Mass of water = 0.415 Kg
Molality of NaBr =?
Molality = mole / mass of water in Kg
Molality of NaBr = 0.142 / 0.415
Molality of NaBr = 0.342 m
Therefore, the molality of NaBr solution is 0.342 m.
Answer:
it's a covalent compound
Explanation:
Two non metals are bonded to together