The pH of a buffer solution : 4.3
<h3>Further explanation</h3>
Given
0.2 mole HCNO
0.8 mole NaCNO
1 L solution
Required
pH buffer
Solution
Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.
![\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}](https://tex.z-dn.net/?f=%5Ctt%20%5Cdisplaystyle%20%5BH%5E%2B%5D%3DKa%5Ctimes%5Cfrac%7Bmole%5C%3Aweak%5C%3Aacid%7D%7Bmole%5C%3Asalt%5Ctimes%20valence%7D)
valence according to the amount of salt anion
Input the value :
![\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3](https://tex.z-dn.net/?f=%5Ctt%20%5Cdisplaystyle%20%5BH%5E%2B%5D%3D2.10%5E%7B-4%7D%5Ctimes%5Cfrac%7B0.2%7D%7B0.8%5Ctimes%201%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D5%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5CpH%3D5-log~5%5C%5C%5C%5CpH%3D4.3)
Answer:
a)4.51
b) 9.96
Explanation:
Given:
NaOH = 0.112M
H2S03 = 0.112 M
V = 60 ml
H2S03 pKa1= 1.857
pKa2 = 7.172
a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.
Therefore, the half points will also be the middle point.
Solving, we have:
pH = (½)* pKa1 + pKa2
pH = (½) * (1.857 + 7.172)
= 4.51
Thus, pH at first equivalence point is 4.51
b) pH at second equivalence point:
We already know there is a presence of SO3-2, and it ionizes to form
SO3-2 + H2O <>HSO3- + OH-
![Kb = \frac{[ HSO3-][0H-]}{SO3-2}](https://tex.z-dn.net/?f=%20Kb%20%3D%20%5Cfrac%7B%5B%20HSO3-%5D%5B0H-%5D%7D%7BSO3-2%7D)
[HSO3-] = x = [OH-]
mmol of SO3-2 = MV
= 0.112 * 60 = 6.72
We need to find the V of NaOh,
V of NaOh = (2 * mmol)/M
= (2 * 6.72)/0.122
= 120ml
For total V in equivalence point, we have:
60ml + 120ml = 180ml
[S03-2] = 6.72/120
= 0.056 M
Substituting for values gotten in the equation ![Kb=\frac{[HSO3-][OH-]}{[SO3-2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BHSO3-%5D%5BOH-%5D%7D%7B%5BSO3-2%5D%7D%20)
We noe have:
![x = [OH-] = 9.11*10^-^5](https://tex.z-dn.net/?f=x%20%3D%20%5BOH-%5D%20%3D%209.11%2A10%5E-%5E5)
=4.04
pH = 14- pOH
= 14 - 4.04
= 9.96
The pH at second equivalence point is 9.96
Answer: B) The energy of the solid increases, and the particles begin to slide past each other.
Explanation:
The energy of the solid matter is required to be increased so that molecules of the solid matter get separated from each other and an transition from solid matter to liquid matter occurs. The liquid components of the matter will slide past each other as they have obtain energy for fluidity. The liquid iron can be molded into any shape and the energy decreases considerably.
Answer:
To prepare 50L of 32% solution you need: 11L of 30% solution, 22L of 50% solution and 17L of 10% solution.
Explanation:
A 32% solution of acid means 32L of acid per 100L of solution. As the chemist wants to make a solution using twice as much of the 50% solution as of the 30% solution it is possible to write:
2x*50% + x*30% + y*10% = 50L*32%
<em>130x + 10y = 1600 </em><em>(1)</em>
<em>-Where x are volume of 30% solution, 2x volume of 50% solution and y volume of 10% solution-</em>
Also, it is possible to write a formula using the total volume (50L), thus:
<em>2x + x +y = 50L</em>
<em>3x + y = 50L </em><em>(2)</em>
If you replace (2) in (1):
130x + 10(50-3x) = 1600
100x + 500 = 1600
100x = 1100
<em>x = 11L -Volume of 30% solution-</em>
2x = 22L -Volume of 50% solution-
50L - 22L - 11L = 17 L -Volume of 10% solution-
I hope it helps!
Answer:
2.49*10⁻¹² mol
Explanation:
Use Avogadro's Number for this equation (6.022*10²³). Divide Avogrado's by the number of atoms you have to find moles. You are answer should be 2.49*10⁻¹² mol.
