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Nadusha1986 [10]
3 years ago
5

If a sample of sodium chloride with a mass of

Chemistry
1 answer:
alex41 [277]3 years ago
6 0
Original molarity was 1.7 moles of NaCl

Final molarity was 0.36 moles of NaCl

Given Information:

Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity

Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity

1. Solve for the molarity of the original (concentrated) solution.

Molarity (M) = moles of solute (mol) / liters of solution (L)

Convert the given information to the appropriate units before plugging in and solving for molarity.

Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)

2. Solve for the molarity of the final (diluted) solution.

Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.

Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution

Molarity (M) of the final solution = 0.36 M NaCl

I hope this helped:))
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Calculate the concentrations of all species present in a 0.26 M solution of ethylammonium chloride (C2H5NH3Cl).
Alina [70]

Answer:

0.00000223

Explanation:

pKa for C2H5NH3+ = 10.7

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pKa + pKb = pKw

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pKb = 14.0 - 10.7

pKb = 3.30

C2H5NH3Cl is a salt of ethylamine and HCl so it will dissolve in water to produce  C2H5NH3^+ + Cl^-

The base hydrolysis reaction:  C2H5NH3^+(aq) + H2O(l) <=> C2H5NH2(aq) + H3O^+(aq)

This reaction is described by Kb.

Kb = [C2H5NH2][H3O^+]/[C2H5NH3^+]

Let [C2H5NH2] = [H3O^+] = x,

so [C2H5NH3^+] = 0.26 - x

Kb = x^2/(0.26 - x) = 2.00 x 10^-11  

Let's solve for x. In this equation,  It is possible to solve without the use quadratic equation. So we can assume that 0.26 - x  is approximately equal to 0.26.  We won't know until we do the calculation.

We get:  x^2 + 2.00 x 10^-11x - 4.99 x 10^-12 = 0

With the use of a quadratic calculator.

x = 2.23 x 10^-6 M = [C2H5NH2] = [H3O^+]

0.26 - x  is just 0.26 M in this problem because 2.23 x 10^-6 M is insignificant.

[C2H5NH3^+] = 0.26 M = [Cl^-]

NOTE:

pH = -log [H3O^+] = -log(2.23 x 10^-6) = 5.65

Ka is the acid dissociation constant

Kb is the base dissociation constant

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3 years ago
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