For parts i - x circle the correct answer. Only one answer per problem. No parti
ID: 973027 • Letter: F
Question
For parts i - x circle the correct answer. Only one answer per problem. No partial credit. Consider the titration of 50.0 mL of 0.100 M HCl with 0.100 M NaOH at 25 degree C. What will the pH be after a total of 40.0 mL of NaOH have been added? 1.255 1.602 1.000 1.954 1.699 Consider the titration of 20.00 mL of 0.100 M HBr with 0.150 M KOH at 25 degree C. What would be the pH of the solution when 20.00 mL of KOH have been added? 12.699 1.602 1.301 12.398 7.000 Consider the titration of 50.00 mL of 0.125 M NH_3, (K_b(NH_3) = 1.8 Times 10^-5) with 0.222 M HCl at 25 degree C. What will the pH be at the midpoint of the titration? 6.54 1.31 9.26 7.00 4.74 For the reaction below, which is first order in A, which of the following statements is false? A rightarrow B The rate constant, k, changes with a change in the concentration of A. The late constant, k, changes with a change in temperature. The rate of the reaction changes with a change in the concentration of A. The rate of the reaction decreases with time. The rate of the reaction changes with a change in temperature.Explanation / Answer
For i)
Moles of HCl = 0.050 L * 0.1 M = 0.005 moles
Moles of NaOH = 0.040 L * 0.1 M = 0.004 moles
Moles of HCl remaining = 0.001 moles
Total Volume = 0.09 L
New molarity of HCL = 0.001 moles / 0.09 L = 0.0111 M
pH = -log(0.01111) = 1.9542, letter d)
For ii)
Moles of HBr = 0.020 L * 0.1 M = 0.002 moles
Moles of KOH = 0.020 L * 0.15 M = 0.003 moles
Moles remaining of KOH = 0.001 moles
New volume = 0.04 L
New concentration of KOH = 0.001 mol / 0.04 L = 0.025 M
pOH = -log(0.025) = 1.602
pH = 14 - 1.602 = 12.398
For iii)
At midpoint, pOH = pKb, so:
pOH = -log(1.8 x 10-5) = 4.74
Hence, pH will be
pH = 14 - 4.74 = 9.26
For iv):
The false statement is the first one, as the rate constant only depends on temperature.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.