The out side t layer is the linnets layer in the earth
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<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine,
, the reaction will be
and we know, pH = -log[H+] and pOH = -log[OH-]
And![kb = \frac{( [( C_{2}H_{5})_{3}NH^{+} ]* OH^{-} )}{[( C_{2}H_{5})_{3}N]}](https://tex.z-dn.net/?f=kb%20%3D%20%20%5Cfrac%7B%28%20%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DNH%5E%7B%2B%7D%20%5D%2A%20%20OH%5E%7B-%7D%20%29%7D%7B%5B%28%20C_%7B2%7DH_%7B5%7D%29_%7B3%7DN%5D%7D%20)
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7
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For triehtylamine,
and we know, pH = -log[H+] and pOH = -log[OH-]
Also, pOH + pH = 14
Now, the Kb value = 5.3 x 10^-4And
thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050
=0.00516 M
Thus, pOH = 2.30
pH = 14 - pOH = 11.7
16.4 grams is the mass of solute in a 500 mL solution of 0.200 M .
sodium phosphate
Explanation:
Given data about sodium phosphate
atomic mass of Na3PO4 = 164 grams/mole
volume of the solution = 500 ml or 0.5 litres
molarity of sodium phosphate solution = 0.200 M
The formula for molarity will be used here to know the mass dissolved in the given volume of the solution:
The formula is
molarity =
putting the values in the equation, we get
molarity x volume = number of moles
0.200 X 0.5= number of moles
number of moles = 0.1 moles
Atomic mass x number of moles = mass
putting the values in the above equation
164 x 0.1 = 16.4 grams
16.4 grams of sodium phosphate is present in 0.5 L of the solution to make a 0.2 M solution.