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Nady [450]
3 years ago
7

Calculate the pH of a solution that is 0.210 M in nitrous acid (HNO2) and 0.290 M in potassium nitrite (KNO2). The acid dissocia

tion constant of nitrous acid is 4.50 × 10-4.
Chemistry
1 answer:
jeka943 years ago
6 0

Answer:

pH = 3.49

Explanation:

We have a buffer system formed by a weak acid (HNO₂) and its conjugate base (NO₂⁻ coming from KNO₂). We can calculate the pH  of a buffer ssytem using the Henderson-Hasselbach equation.

pH = pKa + log [base] / [acid]

pH = -log Ka + log [NO₂⁻] / [HNO₂]

pH = -log 4.50 × 10⁻⁴ + log 0.290 M / 0.210 M

pH = 3.49

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He following balanced equation shows the formation of water. 2H2 + O2 mc020-1.jpg 2H2O How many moles of oxygen (O2) are require
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Answer : The correct option is, 13.7 mole

Solution : Given,

Moles of H_2 = 27.4 moles

The given balanced chemical reaction is,

2H_2+O_2\rightarrow 2H_2O

From the balanced chemical reaction, we conclude that

As, 2 moles of H_2 react with 1 moles of O_2

So, 27.4 moles of H_2 react with \frac{27.4}{2}=13.7 moles of O_2

Therefore, the number of moles of oxygen O_2 required are, 13.7 moles

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Scientists often work on projects for a long time and fail to see sources of error in their research. Which process allows an ou
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Answer:

The correct approach is Option B (Peer Review).

Explanation:

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Other approaches do not apply to the example mentioned. Although the one mentioned is right.

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Which pair of ions can form an ionic bond with each other and why? A.Cu and Ag; They are both metal ions.B. S and O; They have l
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3 years ago
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6. An environmental chemist needs a carbonate buffer of pH 10.00 to study the effects of acid rain on limestone-rich soils. To p
pishuonlain [190]

Answer:

[Na₂CO₃] = 0.094M

Explanation:

Based on the reaction:

HCO₃⁻(aq) + H₂O(l) ↔ CO₃²⁻(aq) + H₃O⁺(aq)

It is possible to find pH using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA]

Where [A⁻] is concentration of conjugate base,  [CO₃²⁻] = [Na₂CO₃] and  [HA] is concentration of weak acid, [NaHCO₃] = 0.20M.

pH is desire pH and pKa (<em>10.00</em>) is -log pka = -log 4.7x10⁻¹¹ = <em>10.33</em>

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Replacing these values:

10.00 = 10.33 + log₁₀ [Na₂CO₃] / [0.20]

<em> [Na₂CO₃] = 0.094M</em>

<em />

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