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Mekhanik [1.2K]
4 years ago
9

What is the ph of a 0.010 m triethanolammonium chloride, (hoc2h2)3nhcl, solution?

Chemistry
1 answer:
azamat4 years ago
8 0

The answer is 4.89


The explanation:


when the Kb of ((HOC2H2)3N) = 5.9 x 10^-7


1- we can get the value of Ka from Kb:


when Ka = Kw / Kb


= 1 x 10^-14 / 5.9 x 10^-7


= 1.69 x 10^-8


2- now we will calculate the [H+] value:


when [H+] = √(Ka*[HA])


=√((1.69E-8)(0.010))


= 1.3 x 10^-5 M


3- the final step we will calculate the PH value from the value of [H+]:


PH = - ㏒ [H+]


= - ㏒ 1.3 x 10^-5


= 4.89

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The reaction 2HI → H2 + I2 is second order in [HI] and second order overall. The rate constant of the reaction at 700°C is 1.57
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Answer:

1.135 M.

Explanation:

  • For the reaction: <em>2HI → H₂ + I₂,</em>

The reaction is a second order reaction of HI,so the rate law of the reaction is: Rate = k[HI]².

  • To solve this problem, we can use the integral law of second-order reactions:

<em>1/[A] = kt + 1/[A₀],</em>

where, k is the reate constant of the reaction (k = 1.57 x 10⁻⁵ M⁻¹s⁻¹),

t is the time of the reaction (t = 8 hours x 60 x 60 = 28800 s),

[A₀] is the initial concentration of HI ([A₀] = ?? M).

[A] is the remaining concentration of HI after hours ([A₀] = 0.75 M).

∵ 1/[A] = kt + 1/[A₀],

∴ 1/[A₀] = 1/[A] - kt

∴ 1/[A₀] = [1/(0.75 M)] - (1.57 x 10⁻⁵ M⁻¹s⁻¹)(28800 s) = 1.333 M⁻¹ - 0.4522 M⁻¹ = 0.8808 M⁻¹.

∴ [A₀] = 1/(0.0.8808 M⁻¹) = 1.135 M.

<em>So, the concentration of HI 8 hours earlier = 1.135 M.</em>

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