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2 years ago

Calculate the pH of a 0.0350M

Chemistry

2 answers:

Triss [41]2 years ago

6 0

Answer:

pH = 12.544

Explanation:

NaOH → Na+ + OH-

0.0350M 0.0350M 0.0350M.........equilibrium

∴ pOH = - log [OH-]

∴ pH + pOH = 14

⇒ [OH-] = 0.0350 M

⇒ pOH = - Log (0.0350)

⇒ pOH = 1.456

⇒ pH = 14 - 1.456

⇒ pH = 12.544

BARSIC [14]2 years ago

4 0

Answer:

12.54

Explanation:

The dissociation equation of NaOH is given below:

NaOH <==> Na+ + OH-

Molarity of NaOH = 0.0350M

From the equation,

Since 1M of NaOH produced 1M of OH-, then, 0.035M of NaOH will also produce 0.035M of OH-..

From the above,

[OH-] = 0.035M

Now, we can solve the pH by doing the following:

pOH = — log [OH-]

pOH = — log 0.035

pOH = 1.46

Recall

pH + pOH = 14

pH = 14 — pOH

pH = 14 — 1.46

pH = 12.54

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A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$