ing ity of St.Francis CHEM 122-Springs GRUENSAUMAclivities and Due DatesExam 2 r
ID: 1029450 • Letter: I
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ing ity of St.Francis CHEM 122-Springs GRUENSAUMAclivities and Due DatesExam 2 review (extra credit)-Due Tuesday, March 20 3/20/2018 10:10 AM/4/ns oe 37 PM Print l Calculator Perlodic Table Question 62 of 74 Map Determine the pH of a solution when 23.7 mL of 0.026 M HNOj is mixed with 19.9 mL of A) 0.0130 M NaOH Number pH = B) distilled water. Number C) 0.0070 M HCI. Number PH- D) 0.156 M KOH. Number PH- O Previous Give Up & View Solution O Check Answer Next 41 Exit Hint about s caroersprivecy pokyerms of use contact search 0Explanation / Answer
pH after addition of 19.9 mL, 0.0130M of NaOH:
Milli-equivalents of HNO3 = 23.7 x 0.026 = 0.616
Milli-equivalents of NaOH = 19.9 x 0.0130 = 0.258
Now, Milli-equivalents of HNO3 in the resultant mixture = 0.616 - 0.258 = 0.358 (ie. acidic solution)
Volume of solution = 23.7+19.9 = 43.6mL
Now, HNO3 molarity or [HNO3] = [H+] = 0.358 / 43.6 = 0.0082M
pH = -log[H+]
pH = -log[0.0082]
pH = 2.08
pH after addition of 19.9 mL of H2O which is neutral:
Milli-equivalents of HNO3 = 23.7 x 0.026 = 0.616
Volume of solution = 23.7+19.9 = 43.6mL
So, HNO3 molarity or [HNO3] = [H+] = 0.616 / 43.6 = 0.0141M
pH = -log[H+]
pH = -log[0.0141]
pH = 1.85
pH after addition of 19.9 mL of 0.007MHCl:
Milli-equivalents of HNO3 = 23.7 x 0.026 = 0.616
Milli-equivalents of HCl = 19.9 x 0.007 = 0.139
Now, Milli-equivalents of HNO3+HCl in the resultant mixture = 0.616 + 0.139 = 0.775 (ie. acidic solution)
Volume of solution = 23.7+19.9 = 43.6mL
Now, net [H+] = 0.775 / 43.6 = 0.0173M
pH = -log[H+]
pH = -log[0.0173]
pH = 1.76
pH after addition of 19.9 mL of 0.156MKOH:
Milli-equivalents of HNO3 = 23.7 x 0.026 = 0.616
Milli-equivalents of KOH = 19.9 x 0.156 = 3.10
Now, Milli-equivalents of NaOH in the resultant mixture = 3.10 - 0.616 = 2.48 (ie. basic solution)
Volume of solution = 23.7+19.9 = 43.6mL
So, NaOH molarity or [NaOH] = [OH-] = 2.48 / 43.6 = 0.0568M
pOH = -log[OH-]
pOH = -log[0.0568]
pOH = 1.24
Now, pH = 14 - pOH
pH = 14 - 1.24
pH = 12.76
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