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kvv77 [185]
2 years ago
12

A scientist has 500 mL of a 2.1 M stock solution. She dilutes the solution, and the volume of the solution after the dilution is

3.25 L. What is the molarity (M) of the diluted solution?
Chemistry
1 answer:
Liono4ka [1.6K]2 years ago
4 0

The molarity of the diluted solution is 0.32 M

Considering the question given above, the following data were obtained:

Volume of stock solution (V₁) = 500 mL

Molarity of stock solution (M₁) = 2.1 M

Volume of diluted solution (V₂) = 3.25 L = 3.25 × 1000 = 3250 mL

<h3>Molarity of diluted solution (M₂) =....? </h3>

The molarity of the diluted solution can be obtained as follow:

<h3>M₁V₁ = M₂V₂</h3>

2.1 × 500 = M₂ × 3250

1050 = M₂ × 3250

<h3>Divide both side by 3250</h3><h3 />

M₂ = 1050 / 3250

<h3>M₂ = 0.32 M</h3>

Therefore, the molarity of the diluted solution is 0.32 M

Learn more: brainly.com/question/22325751

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determine the ph of a buffer that is 0.55 M HNO2 and 0.75 M KNO2. tha value of Ka for HNO2 is 6.8*10^-4
Mariana [72]

Answer:

pH = 3.3

Explanation:

Buffer solutions minimize changes in pH when quantities of acid or base are added into the mix. The typical buffer composition is a weak electrolyte (wk acid or weak base) plus the salt of the weak electrolyte. On addition of acid or base to the buffer solution, the solution chemistry functions to remove the acid or base by reacting with the components of the buffer to shift the equilibrium of the weak electrolyte left or right to remove the excess hydronium ions or hydroxide ions is a way that results in very little change in pH of the system. One should note that buffer solutions do not prevent changes in pH but minimize changes in pH. If enough acid or base is added the buffer chemistry can be destroyed.

In this problem, the weak electrolyte is HNO₂(aq) and the salt is KNO₂(aq). In equation, the buffer solution is 0.55M HNO₂ ⇄ H⁺ + 0.75M KNO₂⁻ . The potassium ion is a spectator ion and does not enter into determination of the pH of the solution. The object is to determine the hydronium ion concentration (H⁺) and apply to the expression pH = -log[H⁺].

Solution using the I.C.E. table:

              HNO₂ ⇄    H⁺   +   KNO₂⁻

C(i)        0.55M       0M      0.75M

ΔC            -x            +x          +x

C(eq)  0.55M - x       x     0.75M + x    b/c [HNO₂] / Ka > 100, the x can be                                    

                                                             dropped giving ...

           ≅0.55M        x       ≅0.75M        

Ka = [H⁺][NO₂⁻]/[HNO₂] => [H⁺] = Ka · [HNO₂]/[NO₂⁻]

=> [H⁺] = 6.80x010⁻⁴(0.55) / (0.75) = 4.99 x 10⁻⁴M

pH = -log[H⁺] = -log(4.99 x 10⁻⁴) -(-3.3) = 3.3

Solution using the Henderson-Hasselbalch Equation:

pH = pKa + log[Base]/[Acid] = -log(Ka) + log[Base]/[Acid]

= -log(6.8 x 10⁻⁴) + log[(0.75M)/(0.55M)]

= -(-3.17) + 0.14 = 3.17 + 0.14 = 3.31 ≅ 3.3

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3 years ago
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Georgia [21]

Answer:

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Explanation:

Please see attached picture for full solution.

3 0
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