Question

What is the pH of a HCOOH solution that is 0.018m?

Answer 1

`pH\text{ = }2.75`

Step-by-step 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:

`HCOOH_((aq))+H_2O_((l))\rightleftarrows HCOO^-_((aq))+H_3O^+_((aq))`

The Pka value for formic acid is 3.75

We have it that:

`\begin{gathered} K_a=10^(-pKa) \\ K_a=10^(-3.75)\text{ = }1.78\text{ }*10^(-4) \end{gathered}`

Using the ICE equation for the equilibrium constant,we have it that:

`K_a\text{ = }(\lbrack HCOO^-\rbrack\lbrack H_3O^+\rbrack)/(\lbrack HCOOH\rbrack\rbrack)`

Thus, we have:

`0.000178=\frac{x^2}{0.018\text{ - x}}^{}`

Since x is very small as opposed to tthe initial acid concentration, we approximate x as 0 below

Thus,we have:

`\begin{gathered} (0.018*0.000178)=x^2 \\ x\text{ = }\sqrt[]{0.018*0.000178} \\ x\text{ = 0.00179} \end{gathered}`

Finally, we know that the pH is the negative logarithm to base 10 of the concentration of hydroxonium ion

`\begin{gathered} pH=-log\lbrack H_3O^+\rbrack_{} \\ pH\text{ = -log\lbrack{}0.00179\rbrack} \\ pH\text{ = 2.75} \end{gathered}`