What causes weather?
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Answer:
- According to the law <br> Mass of reactants = mass of product, here <br> `underset(10 g)(CaCO_(3))rarr underset(4.4 g)(CO_(2))+underset(x)(CaO)` <br> Hence, x = 10 g - 4.4 g = 5.6 g <br> Which is mass of CaO.d
- In the first compound <br> Hydrogen = 5.93 % <br> Oxygen = `(100-5.93)% = 94.07 %` <br> In the second compound <br> Hydrogen = 11.2 % <br> Oxygen `= (100-11.2)%=88.8%` <br> In the first compound the number of parts by mass of oxygen that combine with one part by mass of hydrogen `=(94.07)/(5.93)=15.86` parts ...
- (The ratio of Cu combining with fixed weight of oxygen in black and red oxide is 1 : 2 respectively. Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.) {Check something more in the above attachment!}
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Answer:
pH of buffer after addition of 1 mL of 0,1 M HCl = 7,0
Explanation:
It is possible to use Henderson–Hasselbalch equation to estimate pH in a buffer solution:
pH = pka + log₁₀
Where A⁻ is conjugate base and HA is conjugate acid
The equilibrium of phosphate buffer is:
H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺ Kₐ₂ = 6,20x10⁻⁸; pka=7,2
Thus, Henderson–Hasselbalch equation for 7,00 phosphate buffer is:
7,0 = 7,2 + log₁₀
Ratio obtained is:
0,63 =
As the problem said you can assume [H₂PO₄⁻] = 0,1 M and [HPO4²⁻] = 0,063M
As the amount added of HCl is 0,001 M the concentrations in equilibrium are:
H₂PO₄⁻ ⇄ HPO4²⁻ + H⁺
0,1 M +x 0,063M -x 0,001M -x -<em>because the addition of H⁺ displaces the equilibrium to the left-</em>
Knowing the equation of equilibrium is:
Replacing:
6,20x10⁻⁸ =
You will obtain:
x² -0,064 x + 6,29938x10⁻⁵ = 0
Thus:
x = 0,063 → No physical sense
x = 0,00099990
Thus, [H⁺] in equilibrium is:
0,001 M - 0,00099990 = 1x10⁻⁷
Thus, pH of buffer after addition of 1 mL of 0,1 M HCl =
-log₁₀ [1x10⁻⁷] = 7,0
A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. In this example you can see its effect!
I hope it helps!