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enyata [817]
3 years ago
15

A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s

olution after the addition of 54.0 mL of the KOH solution. Maleic acid has pKa values of 1.92 and 6.27. pH= At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated as H2M, HM−, and M2−, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively. [M2−]= M [HM−]= M [H2M]=
Chemistry
1 answer:
Lilit [14]3 years ago
3 0

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.

           

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(d). Significantly Smaller value

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This is quite a dicey  question to analyze, let us not get too carried away with the simplicity of the mathematical signs involved.

Given that 7.02 × 10⁻² is the number to be compared with.

(a).  7.02 × 10⁻² + 6.10 × 10⁵

from this, we can see a very large number being added to the base number involved, although the base number is carrying a negative power, it does not affect the other greatly.

ANS: the sum would give a significantly larger value

(b). 7.02 × 10⁻² - 6.10 × 10⁵

Here, we can see a very large number subtracted from the base number having a negative power.

the subtraction of this numbers gives a Significantly smaller value than the base number i.e. 7.02 × 10⁻²

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(c). 7.02 × 10⁻² × 6.10 × 10⁵

we would solve this from our basic knowledge of indices

where ⇒ (10ᵃ × 10ᵇ = 10ᵃ⁺ᵇ)

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7.02 × 10⁻² × 6.10 × 10⁵ = (7.02 × 6.10) × 10⁻²⁺⁵ = (7.02 × 6.10) × 10³

this final value (multiplication value) gives a Significantly larger value.

ANS: Significantly larger value

(d). 7.02 × 10⁻² ÷ 6.10 × 10⁵

Also, we apply the indices rule for Division

where ⇒ (10ᵃ ÷ 10ᵇ = 10ᵃ⁻ᵇ)

i.e. 7.02 × 10⁻² ÷ 6.10 × 10⁵ = (7.02 × 6.10) × 10⁻²⁻⁺⁵ = (7.02 × 6.10) × 10⁻⁷

this final value gives a Significantly Smaller value

ANS: Significantly Smaller value

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