Respuesta :
H = 2.7, therefore amount of H+ needed is 10^-2.7 M
u haf 80 ml of hcl and 90 ml of naoh left, therefore 20 ml of hcl used, and 10ml of naoh used.
mols of H+ = 0.02 x 7 x 10^-2 = 1.4 x 10^-3
mols of OH- = 0.01 x 5 x 10^-2 = 5 x 10^-4
H+ and OH- neutralise each other, so remaining mols of H+ = 9 x 10^-4
u need 10^-2.7 mols of H+, so 10^-2.7 - 9 x 10^-4 = 0.001095 mol
vol of HCL needed = 0.001095 / 7 x 10^-2 = 0.0156 L = 15.6 mL
u haf 80 ml of hcl and 90 ml of naoh left, therefore 20 ml of hcl used, and 10ml of naoh used.
mols of H+ = 0.02 x 7 x 10^-2 = 1.4 x 10^-3
mols of OH- = 0.01 x 5 x 10^-2 = 5 x 10^-4
H+ and OH- neutralise each other, so remaining mols of H+ = 9 x 10^-4
u need 10^-2.7 mols of H+, so 10^-2.7 - 9 x 10^-4 = 0.001095 mol
vol of HCL needed = 0.001095 / 7 x 10^-2 = 0.0156 L = 15.6 mL
The given solution has the presence of 100 mL HCl and NaOH. The solution mixing results in the change in pH and to achieve the pH of 2.50, 42.9 mL of HCl is added.
What is neutralization?
The reaction of the acid and the base for the formation of salt and water is termed neutralization.
The given reaction mixture has,
100 mL HCl=0.06 M
100 mL NaOH = 0.05 M
The amount of HCl and NaOH left after mixing was;
HCl = 81 mL
NaOH = 89 mL
The volume of HCl and NaOH in the resulting solution is:
HCl = 19 mL
NaOH = 11 mL
The moles of hydrogen ion in the solution is given as:
[tex]\rm Moles=Molarity\;\times\;Volume\\Moles\;H^+=0.06\;M\;\times\;\dfrac{19\;mL}{1000}\\ Moles\;H^+=0.00114\;mol[/tex]
The moles of hydroxide ion in the solution is given as:
[tex]\rm Moles=Molarity\;\times\;Volume\\Moles\;OH^-=0.05\;M\;\times\;\dfrac{11\;mL}{1000}\\ Moles\;OH^-=0.00055\;mol[/tex]
The neutralization of hydrogen and hydroxide ions is in the 1:1 ratio. The moles of hydrogen left in the reaction is:
[tex]\rm Moles\;H^+\;left=Moles\;H^+-Moles\;OH^-\\Moles\;H^+\;left=0.00114-0.000055\\Moles\;H^+\;left=0.00590\;mol[/tex]
The molarity of the final solution is:
[tex]\rm Molarity=\dfrac{Moles}{Volume}\\\\ Molarity=\dfrac{0.0059}{\frac{11+19\;ml}{1000} }\\\\ Molarity = 0.01967\;M[/tex]
The pH of the solution is given as:
[tex]\rm pH=-logH^+\\pH=-log(0.01967)\\pH=1.71[/tex]
The required hydrogen ion concentration at 2.50 pH is:
[tex]\rm H^+=10^{-pH}\\H^+=10^{-2.50}\\H^+=0.003162\;M[/tex]
The mole equivalent molarity as the final volume is 1 L. The moles of Hydrogen ions required to add is:
[tex]\rm H^+\;required=final-initial\\H^+\;required=0.003162-0.000590\\H^+\;required=0.002572 \;moles[/tex]
The volume of HCl can be given as:
[tex]\rm M_1V_1=M_2V_2\\0.06\;M\;\times\;V_1=0.002572 M\;\times\;1\;L\\V_1=0.0429\;L[/tex]
Thus, the volume of HCl required to add to the solution is 42.9 mL.
Learn more about pH, here:
https://brainly.com/question/15289741
