The answer is 2 because you have to divide these numbers.<span />
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: The concentration of the OH-, CB = 0.473 M.
Explanation:
The balanced equation of reaction is:
2HCl + Ca(OH)2 ===> CaCl2 + 2H2O
Using titration equation of formula
CAVA/CBVB = NA/NB
Where NA is the number of mole of acid = 2 (from the balanced equation of reaction)
NB is the number of mole of base = 1 (from the balanced equation of reaction)
CA is the concentration of acid = 1M
CB is the concentration of base = to be calculated
VA is the volume of acid = 23.65 ml
VB is the volume of base = 25mL
Substituting
1×23.65/CB×25 = 2/1
Therefore CB =1×23.65×1/25×2
CB = 0.473 M.