Question

This is a question I found online for ph but could not find the answer or working out

Answer 1

`12.58`

Step-by-step explanation:

We start by getting the number of moles of the acid

We can get that by multiplying the molarity by the volume

`(25*0.0566)/(1000)\text{ = 0.001415 mole}`

From here, we can get the number of moles of the hydronium ion before the addition of the base

Since the acid is weak, dissociation will be incomplete

We have the dissociation as:

`HX+H_2O\text{ }\rightleftarrows H_3O^{+\text{ }}+X^-`

We can get the number of moles of the hydronium ion by the following steps:

`Ka\text{ = }(\lbrack H_3O^+\rbrack\lbrack X^-\rbrack)/(\lbrack HX\rbrack)`

We also have it that:

`\begin{gathered} PK_a=-log\lbrack K_a\rbrack_{} \\ 4.3\text{ = }-\log \lbrack K_a\rbrack \\ K_a\text{ = 5.01 }*10^(-5) \end{gathered}`

Using an ICE table, we know that the change in the reactant moles is x

Thus, we have it that:

`\begin{gathered} 5.01\text{ }*\text{ }10^(-5)\text{ = }(x* x)/(0.001415-x) \\ x\text{ is small, so we can represent as 0} \\ x^2\text{ = 0.001415}*5.01*10^(-5) \\ x\text{ = 0.0002663 mol} \end{gathered}`

We have the number of moles of hydrogen as x above

Now, we have the pH calculated by subtracting the number of moles of hydroxide ions and dividing by the total volume

We have the hydronium concentration after the addition of the base as follows:

`(0.12(15)-(0.0002663*1000))/((25+15))=0.0383425`

Finally, we can get the pH. What we have from above is pOH since the base is stronger and the acid is weaker. It is expected that what we have is a basic solution.

Mathematically:

`\begin{gathered} pOH=-log\lbrack OH^-\rbrack \\ \text{pOH = -log(0.0383425)} \\ \text{pOH = }1.42 \end{gathered}`
`\begin{gathered} pH\text{ = 14-pOH} \\ pH\text{ = 14-1.42} \\ pH\text{ = 12.58} \end{gathered}`