Answer:
C. CH3COOH, Ka = 1.8 E-5
Explanation:
analyzing the pKa of the given acids:
∴ pKa = - Log Ka
A. pKa = - Log (1.0 E-3 ) = 3
B. pKa = - Log (2.9 E-4) = 3.54
C. pKa = - Log (1.8 E-5) = 4.745
D. pKa = - Log (4.0 E-6) = 5.397
E. pKa = - Log (2.3 E-9) = 8.638
We choose the (C) acid since its pKa close to the expected pH.
⇒ For a buffer solution formed from an acid and its respective salt, we have the equation Henderson-Hausselbach (H-H):
- pH = pKa + Log ([CH3COO-]/[CH3COOH])
∴ pH = 4.5
∴ pKa = 4.745
⇒ 4.5 = 4.745 + Log ([CH3COO-]/[CH3COOH])
⇒ - 0.245 = Log ([CH3COO-]/[CH3COOH])
⇒ 0.5692 = [CH3COO-]/[CH3COOH]
∴ Ka = 1.8 E-5 = ([H3O+].[CH3COO-])/[CH3COOH]
⇒ 1.8 E-5 = [H3O+](0.5692)
⇒ [H3O+] = 3.1623 E-5 M
⇒ pH = - Log ( 3.1623 E-5 ) = 4.5
Answer:
b) The dehydrated sample absorbed moisture after heating
Explanation:
a) Strong initial heating caused some of the hydrate sample to splatter out.
This will result in a higher percent of water than the real one, because you assume in the calculation that the splattered sample was only water (which in not true).
b) The dehydrated sample absorbed moisture after heating.
Usually inorganic salts may absorbed moisture from the atmosphere so this will explain the 13% difference between calculated water percent the real content of water in the hydrate.
c) The amount of the hydrate sample used was too small.
It will create some errors but they do not create a difference of 13% difference as stated in the problem.
d) The crucible was not heated to constant mass before use.
Here the error is small.
e) Excess heating caused the dehydrated sample to decompose.
Usually the inorganic compounds are stable in the temperature range of this kind of experiments. If you have an organic compound which retain water molecules you may decompose the sample forming volatile compounds which will leave crucible so the error will be quite high.
34g C * ( 1 mol / 12.0107 ) * ( 1 mol H2 / 1 mol C ) * ( <span>2.01588 g / 1 mol H2 ) = 5.70657164028741 g H2 = 5.7 g H2
Convert grams of C to moles of C using the given amount of grams and the molar mass ( 12.0107 g/mol ).
Gather the mole ratio from the coefficients in the balanced equation and multiply by the ratio.
Convert moles of H2 to grams of H2 </span> using the given amount of grams and the molar mass ( 2.01588 g/mol )<span>.
Revise your answer to have the correct number of significant figures. </span>
Answer : The freezing point of the solution is, 260.503 K
Solution : Given,
Mass of methanol (solute) = 215 g
Mass of water (solvent) = 1000 g = 1 kg (1 kg = 1000 g)
Freezing depression constant = 
Formula used :
where,
= freezing point of water = 
= freezing point of solution
= freezing point constant
= mass of solute
= mass of solvent
= molar mass of solute
Now put all the given values in the above formula, we get
By rearranging the terms, we get the freezing point of solution.
Therefore, the freezing point of the solution is, 260.503 K
