Question 7 of 13 Map this Faculty and Staff at University of California, Irvine

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Question 7 of 13 Map this Faculty and Staff at University of California, Irvine as been customize sapling learning What is the pH of the solution created by combining 1.00 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 Number after mixin the right into the box below of 0.100 M NaOH with 8.00 mL of 0.100 M HCI- pH after mixing 1.00 mL of Number 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

To do this, you need to calculate the moles of base and HCl in every point, so you'll know what would be in excess and limiting:

a) moles HCl = 0.1 * 0.008 = 0.0008 moles
moles Base added = 0.1 * 0.001 = 0.0001 moles

moles remaining: 0.0008-0.0001 = 0.0007 moles of acid.
C = 0.0007 / 0.009 = 0.078 M
pH = -log(0.078) = 1.11

At this pH the third graph belongs.

b) The acetic acid is a weak acid, and have a pKa = 4.74 and Ka = 1.8x10-5

r: HA + OH- <---------> A- + H2O
i. 0.0008 0.0001 0 0
e: 0.0007 0.0001

pH = pKa + log(A/HA)
pH = 4.74 + log(0.0001/0.0007)
pH = 3.89

At this pH the first graph is the correct.

c) if we diluted the HCl, the concentration is as follow:
C = 0.1 * (8/108) = 0.0074 M

This will be the same concentration for acetic acid. Just work now with these concentrations values, and solve for pH as i did in part a and b, and you will get the answer.

Hope this helps

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