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Sagot :
Answer:
[tex]pH\text{ = }2.75[/tex]Explanation:
Here, we want to get the pH of the given molecule
Firstly,we write the ionization equation so as to get the number of hydrogen atoms from its dissociation
We have this as:
[tex]HCOOH_{(aq)}+H_2O_{(l)}\rightleftarrows HCOO^-_{(aq)}+H_3O^+_{(aq)}[/tex]The Pka value for formic acid is 3.75
We have it that:
[tex]\begin{gathered} K_a=10^{-pKa} \\ K_a=10^{-3.75}\text{ = }1.78\text{ }\times10^{-4} \end{gathered}[/tex]Using the ICE equation for the equilibrium constant,we have it that:
[tex]K_a\text{ = }\frac{\lbrack HCOO^-\rbrack\lbrack H_3O^+\rbrack}{\lbrack HCOOH\rbrack\rbrack}[/tex]Thus, we have:
[tex]0.000178=\frac{x^2}{0.018\text{ - x}}^{}[/tex]Since x is very small as opposed to tthe initial acid concentration, we approximate x as 0 below
Thus,we have:
[tex]\begin{gathered} (0.018\times0.000178)=x^2 \\ x\text{ = }\sqrt[]{0.018\times0.000178} \\ x\text{ = 0.00179} \end{gathered}[/tex]Finally, we know that the pH is the negative logarithm to base 10 of the concentration of hydroxonium ion
[tex]\begin{gathered} pH=-log\lbrack H_3O^+\rbrack_{} \\ pH\text{ = -log\lbrack{}0.00179\rbrack} \\ pH\text{ = 2.75} \end{gathered}[/tex]
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