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mylen [45]
3 years ago
11

What is the pH at the equivalence point in the titration of a 25.7 mL sample of a 0.370 M aqueous nitrous acid solution with a 0

.491 M aqueous potassium hydroxide solution? pH =
Chemistry
1 answer:
expeople1 [14]3 years ago
3 0

Answer:

pH = 8.24

Explanation:

Nitrous acid (HNO₂) reacts with KOH, thus:

HNO₂ + KOH → KNO₂ + H₂O

Moles of HNO₂ are:

0.0257mL ₓ (0.370mol / L) = 0.00951moles.

In equivalence point, the complete moles of nitrous acid reacts with KOH producing potassium nitrite. There are needed:

0.00951mol ₓ (1L / 0.491mol) = 0.01937L ≡ 19.4mL of 0.491M KOH to reach equivalence point.

Total volume in equivalence point is: 19.4mL + 25.7mL = <em>45.1mL</em>

Potassium nitrite is in equilibrium with water, thus:

NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻

Where equilibrium constant, Kb, is defined as:

Kb = 1.41x10⁻¹¹ = \frac{[OH^-][HNO_2]}{[NO_2]}

In equilibrium, molarity of each compound are:

[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X

[HNO₂]: X

[OH⁻]: X

<em>Where X is reaction coordinate</em>

Replacing in Kb:

1.41x10⁻¹¹ = \frac{[X][X]}{[0.211 -X]}

0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²

Solving for X:

X = -1.72x10⁻⁶ <em>FALSE ANSWER. There is no negative concentrations.</em>

X = 1.72x10⁻⁶. <em>Right answer.</em>

That means:

[OH⁻]: 1.72x10⁻⁶M

As pOH is -log [OH⁻] and pH = 14-pOH:

pOH = 5.76; <em>pH = 8.24</em>

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Which of the following represents the least number of molecues?
Mars2501 [29]

Answer:

Answer: a) 20g of H2O (18.02 g/mol) molecules=6.68x10^23

Explanation:

In order to find the amount of molecules of each of the options, we need to follow the following equation.

molecules=\frac{mass(g)x6.022x10^{23}(molecules/mol) }{atomic weight(g/mol)}

So, let´s get the number of molecules for each of the options.

a) molecules=\frac{20(g)x6.022x10^{23}(molecules/mol) }{18.02(g/mol)}=6.68x10^{23}molecules

b) molecules=\frac{77(g)x6.022x10^{23}(molecules/mol) }{16.06(g/mol)}=2.89x10^{24}molecules

c) molecules=\frac{68(g)x6.022x10^{23}(molecules/mol) }{42.09(g/mol)}=9.73x10^{23}molecules

d) molecules=\frac{100(g)x6.022x10^{23}(molecules/mol) }{44.02(g/mol)}=1.37x10^{24}molecules

d) molecules=\frac{84(g)x6.022x10^{23}(molecules/mol) }{20.01(g/mol)}=2.53x10^{24}molecules

the smalest number is in option a)

Best of luck.

7 0
3 years ago
Write the complete balanced equation for the neutralization reaction that occurs when aqueous hydroiodic acid, HI, and sodium hy
Serhud [2]

Answer:

H+ ( aq ) + HCO3- ( aq ) ------> H2O( l ) + CO2 ( g )

Explanation:

The complete reaction when hydroiodic acid and sodium hydrogen carbonate combine, would be as follows -

HI + NaHCO3 ----> NaI + H2O + CO2

net reaction

H2CO3 is highly unstable, and thus decomposes into the water and carbon dioxide you see present as the reactants. If you didn't know already, H2CO3 is also reffered to as carbonic acid. The rest of the elements present on the reactant side are Iodine and Sodium, which is why they are present on the product side as NaI.

Let me include the " physical states " in this reaction as well -

HI ( aq ) + NaHCO3 ( aq ) ----> NaI ( aq ) + H2O ( l ) + CO2 ( g )

Now the complete ionic equation would simply be each compound present as ions in an aqueous solution, so there is no need for an explanation on this step -

H+ ( aq ) + I- ( aq ) + Na+ ( aq ) + HCO3- ( aq ) -------> Na+ ( aq ) + I- ( aq ) + H2O( l ) + CO2 ( g )

The spectator ions in this reaction are I- and Na+, so canceling them out, you would receive the following net ionic equation -

H+ ( aq ) + HCO3- ( aq ) ------> H2O( l ) + CO2 ( g )

<u><em>Hope that helps!</em></u>

4 0
3 years ago
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ale4655 [162]
Mass is measured in kilograms.
5 0
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Help............???????
balandron [24]

B. .175


Hope it helps!

4 0
3 years ago
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aleksandrvk [35]

Answer:

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4 0
3 years ago
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