Respuesta :
Answer:
pH at equivalence point is 8.52
Explanation:
[tex]HCOOH+NaOH\rightarrow HCOO^{-}Na^{+}+H_{2}O[/tex]
1 mol of HCOOH reacts with 1 mol of NaOH to produce 1 mol of [tex]HCOO^{-}[/tex]
So, moles of NaOH used to reach equivalence point equal to number of moles [tex]HCOO^{-}[/tex] produced at equivalence point.
As density of water is 1g/mL, therefore molarity is equal to molality of an aqueous solution.
So, moles of [tex]HCOO^{-}[/tex] produced = [tex]\frac{29.80\times 0.3567}{1000}moles=0.01063moles[/tex]
Total volume of solution at equivalence point = (25+29.80) mL = 54.80 mL
So, at equivalence point concentration of [tex]HCOO^{-}[/tex] = [tex]\frac{0.01063\times 1000}{54.80}M=0.1940M[/tex]
At equivalence point, pH depends upon hydrolysis of [tex]HCOO^{-}[/tex]. So, we have to construct an ICE table.
[tex]HCOO^{-}+H_{2}O\rightleftharpoons HCOOH+OH^{-}[/tex]
I: 0.1940 0 0
C: -x +x +x
E: 0.1940-x x x
So, [tex]\frac{[HCOOH][OH^{-}]}{[HCOO^{-}]}=K_{b}(HCOO^{-})=\frac{10^{-14}}{Ka(HCOOH)}[/tex]
species inside third bracket represent equilibrium concentrations
So, [tex]\frac{x^{2}}{0.1940-x}=5.56\times 10^{-11}[/tex]
or,[tex]x^{2}+(5.56\times 10^{-11}\times x)-(1.079\times 10^{-11})=0[/tex]
So, [tex]x=\frac{-(5.56\times 10^{-11})+\sqrt{(5.56\times 10^{-11})^{2}+(4\times 1.079\times 10^{-11})}}{2}[/tex]
So, [tex]x=3.285\times 10^{-6}M[/tex]
So, [tex]pH=14-pOH=14+log[OH^{-}]=14+logx=14+log(3.285\times 10^{-6})=8.52[/tex]
The study of chemicals is called chemistry. In chemistry, there are two solutions. These are acid and base.
The correct answer is after solving the question is [tex]3.285*10^{-6}[/tex].
What is titration?
- Titration is a common laboratory method of quantitative chemical analysis to determine the concentration of an identified analyte.
- A reagent, termed the titrant or titrator, is prepared as a standard solution of known concentration and volume.
To solve the equation, we have to find the number of moles of [tex]HCOO-[/tex],
hence, the number of moles is as follows:-
[tex]\frac{29.80*0.3567}{1000}\\\\ =0.01063moles[/tex]
All the data required to solve this question is given in the question.
What is the formula used to solve the question?
- [tex]\frac{[HCOOH][OH]^-}{HCOO^-}[/tex]
All the data is given in the question, placed all the values in the equation mentioned above:-
[tex]\frac{5.56*10^{-11}\sqrt{5.56*10{^-11}^2+(4*1.079*10^{-11})} }{2}[/tex]
After solving the equation, the correct answer is [tex]3.285*10^{-6}[/tex] which is the ph of the weak acid and strong base titration.
Hence, the correct answer is [tex]3.285*10^{-6}[/tex]
For more information about the titration, refer to the link:-https://brainly.com/question/10038290
