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12345 [234]
3 years ago
15

Calculate the dissociation constant of nh4oh(aq) if the degree of dissociation of 0.006 mol/kg solution is 0.053 and the activit

y coefficients of all species equal 1
Chemistry
1 answer:
Anastasy [175]3 years ago
7 0

The dissociation equation will be

                         NH4OH   --->        NH4+                   + OH-

Initial                 0.006                        0                          0

Change         -0.006 X 0.053        +0.006 X 0.053      -0.006 X 0.053

Equlibrium     0.006 -0.006 X 0.053      0.006 X 0.053    0.006 X 0.053

Ka = [NH4+] [ OH-] / [NH4OH] = (0.006 X 0.053)^2 / 0.006 -0.006 X 0.053

Ka = 1.78 X 10^-5

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0.2640 g of sodium oxalate is dissolved in a flask and requires 30.74 mL of potassium
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Moles of potassium permanganate = 0.0008

<h3>Further explanation  </h3>

Titration is a procedure for determining the concentration of a solution by reacting with another solution which is known to be concentrated (usually a standard solution). Determination of the endpoint/equivalence point of the reaction can use indicators according to the appropriate pH range  

Reaction

5Na2C2O4(aq) + 2KMnO4(aq) + 8H2SO4(aq) ---> 2MnSO4(aq) + K2SO4(aq) + 5Na2SO4(aq) +  10CO2(g) + 8H2O(1)

The end point ⇒titrant and analyte moles equal

titrant : potassium  permanganate-KMnO4

analyte : sodium oxalate - Na2C2O4

so moles of KMnO4 = moles of Na2C2O4

moles of Na2C2O4(mass = 0.2640 g, MW=134 g/mol) :

\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{0.264}{134 g/mol}\\\\mol=0.002

From equation, mol ratio  Na2C2O4 : KMnO4 = 5 : 2, so mol KMnO4 :

\tt \dfrac{2}{5}\times 0.002=0.0008

6 0
3 years ago
7. Bananas contain potassium. Which form of potassium is most likely found in bananas or in nature? a. K atom b. K+ cation C. K-
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B. K+cation

Explanation:

Not all that sure but hope it will help.

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