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AlladinOne [14]
3 years ago
10

How many moles of NaCI are present in 0.75L of a 0.55M solution?

Chemistry
1 answer:
kolezko [41]3 years ago
6 0

Answer:

,4125 (or .41 if you go by sig figs)

Explanation:

M = (mol)/(L) is a formula for concentration, but we need to solve for moles, so rewrite it as

mol = (L)(M), where we can simply plug in values as follows:

# of mols = (.75L)(.55M) = .4125mol NaCl

The question has 2 significant figures, so you could also put .41 mol NaCl

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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

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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

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Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

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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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