So in your question that ask to calculate the Ph result of the resulting solution if 26 ml of 0.260 M HCI(aq) is added to the following substance. The the result are the following:
A. The result is pH= 14-pOH
B. There are 10ml of 0.26m HCL excees in this reaction so the answer is log(H)+
Answer:
2g
Explanation:
The balanced equation for the reaction is given below:
2Na + 2H2O —> 2NaOH + H2
Next, we'll determine the mass of the water (H2O) that reacted and the mass of H2 produced from the balanced equation.
This is illustrated below:
Molar Mass of H2O = (2x1) + 16 = 18g/mol
Mass of H2O from the balanced equation = 2 x 18 = 36g
Molar Mass of H2 = 2x1 = 2g/mol
Therefore, from the balanced equation above, 36g of H2O reacted to produce 2g of H2.
From the illustration above, we can see that 36g of water will produce 2g of H2.
Answer:
Pure Water
Explanation:
The common ion effect describes the effect on equilibrium that occurs when a common ion (an ion that is already contained in the solution) is added to a solution. The common ion effect generally decreases solubility of a solute(Khan Academy).
NaCl, AgNO3, KCl, BaCl2 solutions all have a common ion with AgCl. As a result of this, AgCl will be much less soluble in these solvents than it is in pure water.
Therefore, AgCl will have the highest solubility in pure water compared to all the solutions listed above.
Answer:
[CH₃COO⁻] [H⁺] pH
0,1 M 0,0025 M 6,30
0,1 M 0,005 M 6,02
0,1 M 0,01 M 5,70
0,1 M 0,05 M 4,74
0,01 M 0,0025 M 5,22
0,01 M 0,005 M 4,75
0,01 M 0,01 M 3,38
0,01 M 0,05 M 1,40
Explanation:
The equilibrium of sodium acetate is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ Kₐ = 1,8x10⁻⁵
Where [CH₃COO⁻] are 0,1 M and 0,01 M and [H⁺] are 0,0025 M 0,005 M 0,01 M and 0,05 M.
For [CH₃COO⁻]=0,1 M and [H⁺]=0,0025M the concentrations in equilibrium are:
[CH₃COO⁻] = 0,1 M - x
[H⁺] = 0,0025 M - x
[CH₃COOH] = x
The expression for this equilibrium is:
Ka =
Replacing:
1,8x10⁻⁵ =
Thus:
0 = x²-0,102518x +2,5x10⁻⁴
Solving:
x = 0,100 ⇒ No physical sense
x = 0,0024995
Thus, [H⁺] = 0,0025-0,0024995 = 5x10⁻⁷
pH = - log [H⁺] = 6,30
Following the same procedure changing both [CH₃COO⁻] and [H⁺] initial concentrations the obtained pH's are:
[CH₃COO⁻] [H⁺] pH
0,1 M 0,0025 M 6,30
0,1 M 0,005 M 6,02
0,1 M 0,01 M 5,70
0,1 M 0,05 M 4,74
0,01 M 0,0025 M 5,22
0,01 M 0,005 M 4,75
0,01 M 0,01 M 3,38
0,01 M 0,05 M 1,40
I hope it helps!