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Bingel [31]
3 years ago
14

A 110.0 ml sample of 0.20 mhf is titrated with 0.10 mcsoh. determine the ph of the solution after the addition of 440.0 ml of cs

oh. the ka of hf is 3.5×10−4.
Chemistry
1 answer:
Margarita [4]3 years ago
6 0
Given:

Concentration of HF = 0.20 m
Volume of HF = 110 ml
Concentration of CsOH = 0.10 m
Volume of CsOH = 440 ml
Ka (HF) = 3.5 x 10^-4

Balanced Chemical Equation:

HF + CsOH ===> CsF + H2O 

pH of the solution = log ([acid]/[base]) + pKa
pH = log ([0.20*(110/1000)]/[0.10*(440/1000)]) - log(3.5x10^-4)
pH = 3.15<span />
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There are two steps in the usual industrial preparation of acrylic acid, the immediate precursor of several useful plastics. In
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The net change in enthalpy for the formation of one mole of acrylic acid from calcium carbide, water and carbon dioxide is 523.2 kJ.

Explanation:

Step 1:

CaC_2(s) + 2H_2O(g)\rightarrow C_2H_2(g) + Ca(OH)_2(s),\Delta H_1=414.0 kJ...[1]

Step 2 :

6C_2H_2(g) + 3CO_2(g) + 4H_2O(g)\rightarrow 5CH_2CHCO_2H(g) \Delta H_2=132.0kJ..[2]

Adding 6 × [1] and [2]:

6CaC_2(s) + 12H_2O(g)\rightarrow 6C_2H_2(g) + 6Ca(OH)_2(s)

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we get :

6CaC_2(s) + 8H_2O(g)+3CO_2(g)\rightarrow 5CH_2CHCO_2H(g)+ 6Ca(OH)_2(s),\Delta H'=?

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\Delta H'=2,626 kJ

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Energy released on formation of 1 mole of acrylic acid:

\frac{ 2,626 kJ}{5 } = 523.2 kJ

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