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Alex_Xolod [135]

3 years ago

11

A solution contains 10. mmol of H3PO4 and 5.0 mmol of NaH2PO4. How many milliliters of 0.10 M NaOH must be added to reach the se

cond equivalence point of the titration of the H3PO4 with NaOH

1 answer:

IRINA_888 [86]3 years ago

8 0

Explanation:

Reaction equations for the given species is as follows.

$H_{3}PO_{4} + 2NaOH \righleftharpoons Na_{2}HPO_{4} + 2H_{2}O$

$NaH_{2}PO_{4} + NaOH \rightleftharpoons Na_{2}HPO_{4} + H_{2}O$

At the first equivalence point we need 2 × 10 mmol NaOH.

At the second equivalence point we need 5 mmol of NaOH.

Hence, total moles of NaOH required is as follows.

(20 + 5) mmol = 25 mmol

We assume that volume of NaOH required is V.

$25 mmol NaOH \times \frac{10^{-3} mol NaOH}{1 mmol NaOH} \times \frac{1000 ml V}{0.10 \text{mol NaOH}}$

= 250 ml V

Thus, we can conclude that 250 ml of 0.10 M NaOH must be added to reach second equivalence point of the titration of the $H_{3}PO_{4}$ and NaOH.

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A solution is prepared by dissolving 27.0 g of urea [(NH2)2CO], in 150.0 g of water. Calculate the boiling point of the solution

andrew11 [14]

<u>Answer:</u> The boiling point of solution is 101.56°C

<u>Explanation:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and boiling point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Boiling point of pure water = 100°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_b$ = molal boiling point elevation constant = 0.52°C/m.g

$m_{solute}$ = Given mass of solute (urea) = 27.0 g

$M_{solute}$ = Molar mass of solute (urea) = 60 g/mol

$W_{solvent}$ = Mass of solvent (water) = 150.0 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=1\times 0.52^oC/m\times \frac{27\times 1000}{60\times 150}\\\\\text{Boiling point of solution}=101.56^oC$

Hence, the boiling point of solution is 101.56°C

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4 years ago

Determine the number of grams found in a 1.20 X 10^30 sample of carbonate ions.

valentina_108 [34]

Answer:

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4 0

3 years ago

Why does carbon have the ability to form multiple bonds?

stepan [7]

Carbon has the ability to form multiple bonds because it has four valence electrons. Having four valence electrons means that carbon has a lot of space to form bonds with other atoms, or multiple bonds, in order to reach the full octet.

Hope this helps!! :)

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3 years ago

True or false: plants, animals, and the environment are not dependent on each other for the use of carbon and oxygen.

Reil [10]

False; animals breathe in oxygen and they produce carbon dioxide when they breathe out, and plants breathe in that carbon dioxide that the animals produce, and with that the plants create oxygen

They basically just swap oxygen for CO2, or CO2 for oxygen! So the answer is false because they DO depend on each other for the use of Oxygen and Carbon dioxide! Hope this helped :)

7 0

3 years ago

The Ksp of AgI is 8.3× 10–17. You titrate 25.00 mL of 0.08160 M NaI with 0.05190 M AgNO3. Calculate pAg after the following volu

11111nata11111 [884]

Answer:

(a) 13.64; (b) 8.04; (c) 2.25

Explanation:

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

$K_{\text{sp}} = {\text{[Ag$^{+}$][I$^{-}$]} = 8.3\times 10^{-17}$

(a) pAg at 35.10 mL

$\text{Moles of I$^{-}$} = \text{0.02500 L} \times \dfrac{\text{0.08160 mol}}{\text{1 L}} = 2.040 \times 10^{-3}\text{ mol/L }\\\text{Moles of Ag$^{+}$} = \text{0.03510 L} \times \dfrac{\text{0.05190 mol}}{\text{1 L}} = 1.822 \times 10^{-3}\text{ mol/L}$

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

I/mol: 1.822 × 10⁻³ 2.040 × 10⁻³

C/mol: -1.822 × 10⁻³ -1.822 × 10⁻³

E/mol: 0 0.218 × 10⁻³

We have a saturated solution of AgI containing 0.218 × 10⁻³ mol of excess I⁻.

V = 25.00 mL + 35.10 mL = 60.10 mL

$\text{[I$^{-}$]} = \dfrac{0.218 \times 10^{-3}\text{ mol}}{\text{0.0610 L}} = 3.57 \times 10^{-3}\text{ mol/L}\\$

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

E/mol·L⁻¹: s 3.57 × 10⁻³ + s

$K_{\text{sp}} = s(3.57 \times 10^{-3} + s) = 8.3\times 10^{-17}\\$

Check for negligibility:

$\dfrac{3.57 \times 10^{-3}}{8.3\times 10^{-17}} = 4.3 \times 10^{13} > 400\\\\\therefore s \ll 3.63 \times 10^{-3}\\K_{\text{sp}} = s\times 3.63 \times 10^{-3}= 8.3\times 10^{-17}\\\\s = \text{[Ag$^{+}$]} = \dfrac{8.3\times 10^{-17}}{3.63 \times 10^{-3}} =2.29 \times 10^{-14}\\\\\text{pAg} = -\log \left (2.29\times 10^{-14} \right) = \mathbf{13.64}$

(b) At equilibrium

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

E/mol·L⁻¹: s s

$K_{\text{sp}} = s\times s = s^{2} = 8.3\times 10^{-17}\\s = \sqrt{8.3\times 10^{-17}} = 9.11 \times 10^{-9}\\\text{pAg} = -\log \left (9.11 \times 10^{-9} \right) = \mathbf{8.04}$

(c) At 47.10 mL

$\text{Moles of Ag$^{+}$} = \text{0.04710 L} \times \dfrac{\text{0.05190 mol}}{\text{1 L}} = 2.444 \times 10^{-3}\text{ mol}$

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

I/mol: 2.444 × 10⁻³ 2.040 × 10⁻³

C/mol: -2.040 × 10⁻³ -2.040 × 10⁻³

E/mol: 0.404 × 10⁻³ 0

V = 25.00 mL + 47.10 mL = 72.10 mL

$\text{[Ag$^{+}$]} = \dfrac{0.404 \times 10^{-3}\text{ mol}}{\text{0.0721 L}} = 5.61 \times 10^{-3}\text{ mol/L}\\\text{pAg} = -\log(5.61 \times 10^{-3}) = \mathbf{2.25}$

6 0

4 years ago