What coefficients should be added in order to make this a balanced equation?

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

expeople1 [14]

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 = 45.1mL

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

Where X is reaction coordinate

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 FALSE ANSWER. There is no negative concentrations.

X = 1.72x10⁻⁶. Right answer.

That means:

[OH⁻]: 1.72x10⁻⁶M

As pOH is -log [OH⁻] and pH = 14-pOH:

pOH = 5.76; pH = 8.24

3 0

3 years ago

A little boy has placed his pet squirrel into a balloon with a volume of 3L at a pressure of 1.0 atm. If the boy takes his pet s

s2008m [1.1K]

Answer: The squirrel's balloon will be 0.86 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

$P_1=1.0atm\\V_1=3L\\P_2=3.5atm\\V_2=?$

Putting values in above equation, we get:

$1.0\times 3L=3.5\times V_2\\\\V_2=0.86$

Thus the squirrel's balloon will be 0.86 L

7 0

2 years ago

What volume of 0.08892 M HNO3 is required to react completetly with 0.2352 g of potassium hydrogen phosphate?

galina1969 [7]

Answer:

0.0303 Liters

Explanation:

Given:

Mass of the potassium hydrogen phosphate = 0.2352

Molarity of the HNO₃ Solution = 0.08892 M

Now,

From the reaction it can be observed that 1 mol of potassium hydrogen phosphate reacts with 2 mol of HNO₃

The number of moles of 0.2352 g of potassium hydrogen phosphate

= Mass / Molar mass

also,

Molar mass of potassium hydrogen phosphate

= 2 × (39.09) + 1 + 30.97 + 4 × 16 = 174.15 g / mol

Number of moles = 0.2352 / 174.15 = 0.00135 moles

thus,

The number of moles of HNO₃ required for 0.00135 moles

= 2 × 0.00135 mol of HNO₃

= 0.0027 mol of HNO₃

Now,

Molarity = Number of Moles / Volume

thus,

for 0.0027 mol of HNO₃, we have

0.08892 = 0.0027 / Volume

or

Volume = 0.0303 Liters

8 0

3 years ago

Read 2 more answers

You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acetate plus acetate concentration

vovangra [49]

Ans: Acetic acid = 90.3 mM and Sodium acetate = 160 mM

Given:

Acetic Acid/Sodium Acetate buffer of pH = 5.0

Let HA = acetic acid

A- = sodium acetate

Total concentration [HA] + [A-] = 250 mM ------(1)

pKa(acetic acid) = 4.75

Based on Henderson-Hasselbalch equation

pH = pKa + log[A-]/[HA]

[A-]/[HA] = 10^(pH-pKa) = 10^(5-4.75) = 10^0.25 = 1.77

[A-] = 1.77[HA] -----(2)

From (1) and (2)

[HA] + 1.77[HA] = 250 mM

[HA] = 250/2.77 = 90.25 mM

[A-] = 1.77(90.25) = 159.74 mM

7 0

3 years ago

Prepare a 0.50 L of a 1.0 M solution of potassium sulfate. Determine the amount of potassium sulfate needed in correct significa

Lera25 [3.4K]

Answer:

B) 87g

Explanation:

Usatestprep said it

3 0

3 years ago