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Romashka [77]
3 years ago
10

The default solution is water, describe the solution. What is the pH?

Chemistry
2 answers:
Nonamiya [84]3 years ago
8 0
The pH of pure water is 7.
Oksana_A [137]3 years ago
3 0
PH pure water.
Because that’s they only one that’s makes more sense
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Calculate the new volume of 1.23 mL of a gas at 32 C is subjected to drop in temperature of 20 degrees Celsius
kotykmax [81]

Answer:

1,15mL = V₂

Explanation:

Based on Charle's law the volume is directely proportional to the absolute temperature in a gas under constant pressure. The equation is:

V₁T₂ = V₂T₁

<em>Where V is volume and T absolute temperature of a gas where 1 is initial state and 2, final state.</em>

The V₁ is 1.23mL

T₁ = 32°C + 273.15 = 305.15K

T₂ = T₁ - 20°C = 285.15K

Replacing:

1.23mL*285.15K = V₂*305.15K

<h3>1,15mL = V₂</h3>

<em />

8 0
3 years ago
Benzoic acid is a natural fungicide that naturally occurs in many fruits and berries. The sodium salt of benzoic acid, sodium be
AlexFokin [52]

Answer:

a. pH = 2.52

b. pH = 8.67

c. pH = 12.83

Explanation:

The equation of the titration between the benzoic acid and NaOH is:

C₆H₅CO₂H + OH⁻ ⇄ C₆H₅CO₂⁻ + H₂O    (1)

a. To find the pH after the addition of 20.0 mL of NaOH we need to find the number of moles of C₆H₅CO₂H and NaOH:

\eta_{NaOH} = C*V = 0.250 M*0.020 L = 5.00 \cdot 10^{-3} moles

\eta_{C_{6}H_{5}CO_{2}H}i = C*V = 0.300 M*0.050 L = 0.015 moles  

From the reaction between the benzoic acid and NaOH we have the following number of moles of benzoic acid remaining:

\eta_{C_{6}H_{5}CO_{2}H} = \eta_{C_{6}H_{5}CO_{2}H}i - \eta_{NaOH} = 0.015 moles - 5.00 \cdot 10^{-3} moles = 0.01 moles

The concentration of benzoic acid is:

C = \frac{\eta}{V} = \frac{0.01 moles}{(0.020 + 0.050) L} = 0.14 M

Now, from the dissociation equilibrium of benzoic acid we have:

C₆H₅CO₂H + H₂O ⇄ C₆H₅CO₂⁻ + H₃O⁺  

0.14 - x                            x                x

Ka = \frac{[C_{6}H_{5}CO_{2}^{-}][H_{3}O^{+}]}{[C_{6}H_{5}CO_{2}H]}

Ka = \frac{x*x}{0.14 - x}

6.5 \cdot 10^{-5}*(0.14 - x) - x^{2} = 0   (2)  

By solving equation (2) for x we have:          

x = 0.0030 = [C₆H₅CO₂⁻] = [H₃O⁺]

Finally, the pH is:

pH = -log([H_{3}O^{+}]) = -log (0.0030) = 2.52

b. At the equivalence point, the benzoic acid has been converted to its conjugate base for the reaction with NaOH so, the equilibrium equation is:

C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻     (3)

The number of moles of C₆H₅CO₂⁻ is:

\eta_{C_{6}H_{5}CO_{2}^{-}} = \eta_{C_{6}H_{5}CO_{2}H}i = 0.015 moles

The volume of NaOH added is:

V = \frac{\eta}{C} = \frac{0.015 moles}{0.250 M} = 0.060 L

The concentration of C₆H₅CO₂⁻ is:

C = \frac{\eta}{V} = \frac{0.015 moles}{(0.060 L + 0.050 L)} = 0.14 M

From the equilibrium of equation (3) we have:

C₆H₅CO₂⁻ + H₂O  ⇄  C₆H₅CO₂H + OH⁻  

0.14 - x                              x               x

Kb = \frac{[C_{6}H_{5}CO_{2}H][OH^{-}]}{[C_{6}H_{5}CO_{2}^{-}]}

(\frac{Kw}{Ka})*(0.14 - x) - x^{2} = 0

(\frac{1.00 \cdot 10^{-14}}{6.5 \cdot 10^{-5}})*(0.14 - x) - x^{2} = 0

By solving the equation above for x, we have:

x = 4.64x10⁻⁶ = [C₆H₅CO₂H] = [OH⁻]

The pH is:

pOH = -log[OH^{-}] = -log(4.64 \cdot 10^{-6}) = 5.33

pH = 14 - pOH = 14 - 5.33 = 8.67

     

c. To find the pH after the addition of 100 mL of NaOH we need to find the number of moles of NaOH:

\eta_{NaOH}i = C*V = 0.250 M*0.100 L = 0.025 moles

From the reaction between the benzoic acid and NaOH we have the following number of moles remaining:                              

\eta_{NaOH} = \eta_{NaOH}i - \eta_{C_{6}H_{5}CO_{2}H} = 0.025 moles - 0.015 moles = 0.010 moles                          

The concentration of NaOH is:

C = \frac{\eta}{V} = \frac{0.010 moles}{0.100 L + 0.050 L} = 0.067 M

Therefore, the pH is given by this excess of NaOH:         

pOH = -log([OH^{-}]) = -log(0.067) = 1.17

pH = 14 - pOH = 12.83

I hope it helps you!    

4 0
3 years ago
Using a chemical equation to find moles of product from moles...
kaheart [24]

Answer: 0.600 moles

Balanced eqn for decomposition of the rocket fuel:

2 NH4ClO4 (s) ----> N2 (g) + Cl2 (g) + 2 O2 (g) + 4 H2O (g)

Since, 2 moles of NH4ClO4 produces 2 moles of O2 gas, this means that 0.6 moles of NH4ClO4 will produce 0.6 moles of O2 gas

7 0
2 years ago
At what temperature does water become a gas?
uranmaximum [27]
The answers is d. 100o c
3 0
4 years ago
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When a substance dissociates into OH- ions, they are classified as ________ ?
djyliett [7]
When a substance dissociates into OH- ions, they are classified as bases.
4 0
3 years ago
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