Question

24. 00 ml of a 0. 25 m naoh solution is titrated with 0. 10m hcl. What is the ph of the solution after 24. 00 ml of the hcl has been added?.

Answer 1

pH of the solution after 24. 00 ml of the hcl has been added is 12.87

millimoles NaOH = mL x M = 24.00 mL x 0.25 M = 6.00

millimoles HCl = 24.00 mL x 0.10 M = 2.40

total volume = 48.00 mL

.................................NaOH + HCl ==>NaCl + H2O

initial.........................6.00.........0............0.........0

added.....................................2.40............................

change.................... -2.40......-2.40.........+2.40.... +2.40

equilibrium.................3.60.........0..............2.40.......2.40

The NaCl contributes nothing to the pH of the final solution. The pH is determined by the excess of NaOH present. (NaOH) = millimoles/mL = 3.60/48.00 = 0.075 M = (OH^-)

pOH = -log (OH^-). Then

pOH = -log (0.075)

pOH =1.1249

As we know,

pH + pOH = pKw = 14.00

pH=14-pOH

pH=14-1.1249

pH=12.87

What is pH?

pH is a logarithmic measure of an aqueous solution's hydrogen ion concentration. pH = -log[H+], where log is the base 10 logarithm and [H+] is the concentration of hydrogen ions in moles per liter.

The pH of an aqueous solution describes how acidic or basic it is, with a pH less than 7 being acidic and a pH greater than 7 being basic. A pH of 7 is regarded as neutral (e.g., pure water). pH values typically range from 0 to 14, though very strong acids may have a negative pH and very strong bases may have a pH greater than 14.

Learn more about pH:

brainly.com/question/491373

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