Question

What is the pH of a solution in which 15 mL of 0.10 M NaOH is added to 25 mL of 0.10 M HCl?

Answer 1

The pH of the solution is 1.60.

Step-by-step explanation:

1st) It is necessary to write and balance the chemical reaction:

`NaOH+HCl\rightarrow NaCl+H_2O`

Now we can see that 1 mole of NaOH reacts with 1 mole of HCl.

2nd) We have to calculate the moles contained in 15mL of 0.10M NaOH solution and the moles contained in 25mL of 0.10M HCl solution:

• Moles contained in NaOH solution:

`\begin{gathered} 1000mL-0.10moles \\ 15mL-x=(15mL*0.10moles)/(1000mL) \\ x=1.5*10^(-3)moles \end{gathered}`

• Moles contained in HCl solution:

`\begin{gathered} 1000mL-0.10moles \\ 25mL-x=(25mL*0.10moles)/(1000mL) \\ x=2.5x10^(-3)moles \end{gathered}`

Now we know that there are 1.5x10^-3 moles of NaOH and 2.5x10^-3 moles of HCl.

3rd) According to the stoichiometry of the reaction, 1 mole of NaOH reacts with 1 mole of HCl, so in this case, 1.5x10^-3 moles of NaOH will react with 1.5x10^-3 moles of HCl, because NaOH will be the limiting reactant and HCl will be the excess reactant.

So, now we have to calculate the excess of HCl:

2.5x10^-3moles - 1.5x10^-3moles = 1x10^-3moles

Now we know that there are 1x10^-3 moles of HCl left.

4th) Excess HCl will remain dissociated into H+ and Cl-, according to the following equation:

`HCl\rightarrow H^++Cl^-`

That means that for every mole of HCl, 1H+ dissociates. So, in this case, there are 1x10^-3 moles of H+.

Remember that these moles are contained in 40mL, so the molarity of H+ is 0.025M:

`\begin{gathered} 40mL-1x10^(-3)moles \\ 1000mL-x=(1000mL*1x10^(-3)moles)/(40mL) \\ x=0.025moles \end{gathered}`

5th) Finally, we can calculate the pH of the solution, by replacing the H+ concentration in the pH formula:

`\begin{gathered} pH=-log\lbrack H^+\rbrack \\ pH=-log\lbrack0.025\rbrack \\ pH=1.60 \end{gathered}`

So, the pH of the solution is 1.60.