Respuesta :
Answer:
The pH of the solution is 1.60.
Explanation:
1st) It is necessary to write and balance the chemical reaction:
[tex]NaOH+HCl\rightarrow NaCl+H_2O[/tex]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:
[tex]\begin{gathered} 1000mL-0.10moles \\ 15mL-x=\frac{15mL*0.10moles}{1000mL} \\ x=1.5*10^{-3}moles \end{gathered}[/tex]• Moles contained in HCl solution:
[tex]\begin{gathered} 1000mL-0.10moles \\ 25mL-x=\frac{25mL*0.10moles}{1000mL} \\ x=2.5x10^{-3}moles \end{gathered}[/tex]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:
[tex]HCl\rightarrow H^++Cl^-[/tex]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:
[tex]\begin{gathered} 40mL-1x10^{-3}moles \\ 1000mL-x=\frac{1000mL*1x10^{-3}moles}{40mL} \\ x=0.025moles \end{gathered}[/tex]5th) Finally, we can calculate the pH of the solution, by replacing the H+ concentration in the pH formula:
[tex]\begin{gathered} pH=-log\lbrack H^+\rbrack \\ pH=-log\lbrack0.025\rbrack \\ pH=1.60 \end{gathered}[/tex]So, the pH of the solution is 1.60.
