Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation

Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.
Answer:
I would say the answer is 2
Answer:
1160mg
Explanation:
Molarity = number of moles ÷ volume
According to the information in the question, molarity = 1.25 M, volume = 18.0 mL = 18/1000 = 0.018L
M = n/V
n = M × V
n = 1.25 × 0.018
n = 0.0225moles.
Using mole = mass/molar mass, to find the mass of lithium phosphide (Li3P)
Molar mass of Li3P = 6.9(3) + 31 = 51.7g/mol
mole = mass/molar mass
0.0225 = mass/51.7
mass = 1.16grams.
In milligrams (mg), mass of Li3P = 1.16 × 1000 = 1160mg
Once molecules are close enough to touch, intermolecular forces become replusive in order to prevent the molecules from overlapping.