Initial [ HCO2H] = moles * volume
=0.35 moles * 1 L = 0.35 M
by using ICE table:
HCO2H ↔ H+ + HCO2-
initial 0.35 M 0 0
change - X +X +X
Equ (0.35 - X) X X
∴ Ka = [H+][HCO2-] / [HCO2H]
by substitution:
1.8 x 10^-4 = X^2 / (0.35-X) by solving for X
∴ X = 0.0079 or 7.9 x 10^-3
∴ [H+] = X = 7.9 x 10^-3 M
<span>E = ħc / λ
ħ = plancks constant = 6.626x10^-34 Js
c = speed of light = 2.999x10^8 m/s
λ = wavelength of light = 670.8x10^-9 m
E = (6.626x10^-34 Js) x (2.999x10^8 m/s) x (1 / 670.8x10^-9 m)
E = 2.962x10^-19 J
</span><span>3x10^8 / (670.8 * 10^-9)
=4.47x10^14 Hz
4.47x10^14 Hz multiplied by plank's constant = 2.9634x10^-19
</span><span>
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Find moles:
<span>36.0 g of glucose divided by 180 g/mol = 0.200 moles of glucose </span>
<span>find molarity: </span>
<span>0.200 moles of glucose / 2 litres = 0.100 Molar solution </span>
Answer : The pH of a 0.1 M phosphate buffer is, 6.86
Explanation : Given,
Concentration of acid = 0.1 M
Concentration of conjugate base (salt) = 0.1 M
Now we have to calculate the pH of buffer.
Using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
Now put all the given values in this expression, we get:
Therefore, the pH of a 0.1 M phosphate buffer is, 6.86
Answer:
Explanation:
F Cl Br belongs to the Same group
