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:
Explanation:
The pressure at the bottom of the tank is:
The force exerted on the circular bottom is:
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Answer:
0.550
Explanation:
The absorbance (A) of a substance depends on its concentration (c) according to Beer-Lambert law.
A = ε . <em>l</em> . c
where,
ε: absorptivity of the species
<em>l</em>: optical path length
A 45 mM phosphate solution (solution A) had an absorbance of 1.012.
A = ε . <em>l</em> . c
1.012 = ε . <em>l</em> . 45 mM
ε . <em>l</em> = 0.022 mM⁻¹
We can find the concentration of the second solution using the dilution rule.
C₁ . V₁ = C₂ . V₂
45mM . 11mL = C₂ . 20.0 mL
C₂ = 25 mM
The absorbance of the second solution is:
A = (ε . <em>l</em> ). c
A = (0.022 mM⁻¹) . 25 mM = 0.55 (rounding off to 3 significant figures = 0.550)
