What is the pH of the solution created by combining 11.00 mL of 0.100 M NaOH wit

Question

What is the pH of the solution created by combining 11.00 mL of 0.100 M NaOH with 8.00 mL of each of the acid solutions? Note: Make simplifying assumptions, do not use the quadratic formula.

Question 8 of 13 this Faculty and Staff at University of California, Irvine as been customize sapling learning What is the pH of the solution created by combining 11.00 mL of 0.100 M NaOH with 8.00 mL of each of the acid solutions? Note: Make simplifying assumptions, do not use the quadratic formula Drag and drop the graph that demonstrates the point calculated to the right into the box below pH after mixing 11.00 mL Number of 0.100 M NaOH with 8.00 mL of 0.100 M HCI- pH after mixing 11.00 mLNumber of 0.100 M NaOH with 8.00 mL of 0.100 M HC2H302-L What are the pH values if you take into account that the 8.00 mL of 0.100 M acid was first diluted with 100 mL water (as it wil be in the experiment you will perform)? Number pH pH pH after neutralizationn of diluted HCl solution Base Volume Base Volume Number pH after neutralization of diluted HC2H302 solution Base Volume Base Volume Previous Give Up & View Solution 0 Check Answer Next Exit Hint

Explanation / Answer

1) pH after mixijg 11 ml of 0.1 M NaOH with 8 ml of 0.1 M HCl

excess [OH-] = 0.1 M x 3 ml/19 ml = 0.0158 M

pOH = -log[OH-] = 1.80

pH = 14 - pOH = 12.2

2) pH after mixing 11.0 ml of 0.1 M NaOH with 8.0 ml of 0.1 M HC2H3O2

excess [OH-] = 0.1 M x 3 ml/19 ml = 0.0158 M

pOH = -log[OH-] = 1.80

pH = 14 - pOH = 12.2

If 8 ml of 0.1 M acid is diluted with 100 ml water

final concentration of acid = 0.1 M x 8 ml/108 ml = 0.0074 M

3) pH after neutralization of diluted HCl

mixing 11 ml of 0.1 M NaOH with 108 ml of 0.0074 M HCl

moles of HCl = 0.0074 M x 108 ml = 0.8 mmol

moles of NaOH = 0.1 M x 11 ml = 1.1 mmol

excess [OH-] = 0.3/119 = 0.0025 M

pOH = -log[OH-] = 2.6

pH = 14 - pOH = 11.4

4) pH after neutralization of diluted HC2H3O2

mixing 11 ml of 0.1 M NaOH with 108 ml of 0.0074 M of HC2H3O2

moles of NaOH = 1.1 mmol

moles of HC2H3O2 = 0.0074 M x 108 ml = 0.8 mmol

excess [OH-] = 0.3 mmol/119 ml = 0.0025 M

pOH = -log[OH-] = 2.6

pH = 14 - pOH = 11.4

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