Question

1. A 100.0-mL sample of 0.500 M sodium hydroxide is titrated with 0.100 M nitric acid. Calculate the pH before the titration begins.

2. A 1.0-L buffer solution contains 0.100 mol HCN and 0.100 mol LiCN. The value of Ka for HCN is 4.9 x 10-10. Because the initial amounts of acid and conjugate base are equal, the pH of the buffer is equal to pKa = -log (4.9 x 10-10) = 9.31. Calculate the new pH after the addition of 0.010 mol of HNO3 to the buffer. Ignore any change of volume brought about by this addition.

3. A 100.0-mL sample of 0.100 M HCHO2 is titrated with 0.200 M NaOH. Calculate the pH after adding 30.00 mL of the base.

Answer #1

**1.**

Before the titration begins, there is only NaOH present in the solution.

NaOH is a strong base. Hence, it completely ionizes in water.

Thus, 0.500 M of NaOH ionizes to give 0.500 M of OH^{-}
in aqueous solution.

Thus,

pOH = - log[OH^{-}]

= - log(0.500)

= 0.301

Now,

pH + pOH = 14

or, pH = 14 - pOH

= 14 - 0.301

= 13.7

Therefore, the pH before the titration begins =
**13.7**

75.0 ml sample of 0.200 M sodium hydroxide is titrated with
.200 M nitric acid. calculate ph
A) after adding 40.00 mL of HNO3
B) at the equivalent point

16.6
A 50.0-mL sample of 0.200 M sodium hydroxide is
titrated with 0.200 M nitric acid.
You may want to reference (Pages 662 - 674) Section 16.4 while
completing this problem.
Part A
Calculate the pH of the solution, after you add a total of 51.9
mL 0.200 M HNO3.
Express your answer using two decimal places.
pH =
nothing

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with
0.500 M HNO3. Calculate the pH after the addition of 25.0 mL of
HNO3.

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5)
is titrated with 0.500 M HNO3. Calculate the pH after the
addition of 15.0 mL of HNO3.

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with
0.500 M HNO3. Calculate the pH after the addition of 13.0 mL of
HNO3.

A 75.0-mL volume of 0.200 M NH3 (Kb=1.8×10−5) is titrated with
0.500 M HNO3. Calculate the pH after the addition of 23.0 mL of
HNO3. Express your answer numerically.

If a sample of sodium hydroxide is titrated with 50.00 mL of
0.400 M nitric acid, how many
grams of the base are in the sample?

1. A 100 mL sample of 0.100 M weak base, B (Kb = 3.7 x 10-4) is
titrated with 0.250 M HNO3. Calculate the pH after the addition of:
a) 0 mL acid b) 20 mL acid

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3.
Determine the pH of the solution after the addition of 100.0 mL of
HNO3. The Kb of NH3 is 1.8 × 10^-5.

1. A25.0 mL sample of 0.100 M lactic acid (hc3h5o5, pka= 3.86)
is titrated with a 0.100 M NaOH solution. Calculate the pH after
the addition of 0.0, 4.0, 8.0, 12.5, 20.0, 24.0, 24.5, 24.9, 25.0,
25.1, 26.0, 28.0, and 30.0 of the NaOH. Plot the results of your
calculation, as a pH versus mililiters of NaOH added.
2. Repeat the procedure in problem 1, but the titration of 25.0
mL 0.100 M NH3 (kb= 1.8*10^-5) with 0.100 M HCl....

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