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ludmilkaskok [199]
3 years ago
7

Calculate the concentration of ammonium nitrate in a solution prepared by dissolving 3.20 g of the salt in enough water to make

100. mL of solution, then diluting 2.00 mL of this solution to a volume of 25.00 mL.
Chemistry
1 answer:
Tpy6a [65]3 years ago
3 0

Answer:

.032 M .

Explanation:

Molecular weight of ammonium nitrate is 80  .

3.2 g = 3.2 / 80 moles

= .04 moles

volume = 100 mL = 0.1 L

Molarity of 100 mL solution = .04 moles / 0.1 L

= 0.4 M solution.

Now 2 mL solution of 0.4 M is diluted to a volume of 25 mL .

Using the formula S₁ V₁ = S₂V₂

0.4 M x 2 mL = S₂ x 25 mL

S₂ = .4 x 2 / 25

= .032 M

Hence required concentration is .032 M .

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pH = 3.49

Explanation:

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What is the Ka of a weak acid (HA) if the initial concentration of weak acid is 4.5 x 10-4 M and the pH is 6.87? (pick one)
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Answer:

Ka = 4.04 \times 10^{-11}

Explanation:

Initial concentration of weak acid = 4.5 \times 10^{-4}\ M

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pH = -log[H^+]

[H^+]=10^{-pH}

[H^+]=10^{-6.87}=1.35 \times 10^{-7}\ M

HA dissociated as:

HA \leftrightharpoons H^+ + A^{-}

(0.00045 - x)    x     x

[HA] at equilibrium = (0.00045 - x) M

x = 1.35 \times 10^{-7}\ M

Ka = \frac{[H^+][A^{-}]}{[HA]}

Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 - 0.000000135}

0.000000135 <<< 0.00045

Therefore, Ka = \frac{(1.35 \times 10^{-7})^2}{0.00045 } = 4.04 \times 10^{-11}

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