Question

A 20.0 mL of 0.20 M HNO_{3} was titrated with 10.0 mL of
0.20 M NaOH. Determine the pH of the solution and record below with
correct significant figures.

Answer #1

Here we are having 20 ml of 0.20 M HNO_{3} .

So Number of moles of H^{+} = Volume * molarity = 20*
10^{-3} l * 0.20 mol l^{-1} = 0.004 moles

Similarly we have 10.0 mL of 0.20 M NaOH .

So Number of moles of OH^{-} = Volume *
molarity = 10* 10^{-3} l * 0.20 mol l^{-1} = 0.002
moles.

Each mole of OH^{-} reacts with one mole of
H^{+} . So all 0.002 moles of OH^{-} will react
with corresponding 0.002 moles of H^{+}.

Number of moles of H^{+} left = 0.004 mol -
0.002 mol = 0.002 mol

Meanwhile Total volume of solution has become 40 ml since we
have mixed 20 ml of HNO_{3} and 20 ml of NaOH.

Hence , Concentration of H^{+} = Moles of H^{+}
/ Volume of solution ( in L ) = 0.002 mol / 0.040 L = 0.05 M

**pH of solution = -log [ H ^{+} ] = - log(0.05) =
1.301**

A 600.0 mL sample of 0.20 M HF is titrated with 0.20 M NaOH.
Determine the pH of the solution after the addition of 100.0 mL of
NaOH. The Ka of HF is 3.5 * 10^-4

If 10.0 mL of 0.20 M NaOH is added to 50.0 mL of 0.10 M HCl,
what will be the pH of the resulting solution?

When 10.0 mL of 0.10 M HCOOH is titrated with .10 M NaOH, at the
equivalence point of the pH the solution will be:
A) Greater than 7 B) 7.00 C) Less than 7

A 25.0 mL sample of a 0.115 M solution of acetic acid is
titrated with a 0.144 M solution of NaOH. Calculate the pH of the
titration mixture after 10.0, 20.0, and 30.0 mL of base have been
added. (The Ka for acetic acid is 1.76 x 10^-5).
10.0 mL of base =
20.0 mL of base =
30.0 mL of base =

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.

Calculate the pH after 10.0 mL of 0.400 M NaOH is added to 20.0
mL of 0.50 M CH3COOH.(Ka of CH3COOH is 1.8 *
10−5)
The answer is supposed to be 4.57, but I'm getting 4.37. I'm so
close but need a little guidance.

A volume of 90.0 mL of a 0.820 M HNO3 solution is
titrated with 0.220 M KOH. Calculate the volume of KOH
required to reach the equivalence point.
Express your answer to three significant figures and include the
appropriate units.

What is the pH of the solution formed after 20.0 mL of 0.20 M
ketamine, C13H16ClNO, is titrated with 5.0 mL of 1.0 M HI? Kb = 3.0
×10–7 for C13H16ClNO
A. 4.60
B. 7.00
✓C. 1.40
D. 2.20

4
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.
8.72
6.58
3.44
10.56
5.28

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.
Please show work. a) 10.56 b) 5.28 c) 6.58 d) 8.72 e) 3.44

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 1 minute ago

asked 9 minutes ago

asked 11 minutes ago

asked 14 minutes ago

asked 14 minutes ago

asked 25 minutes ago

asked 28 minutes ago

asked 28 minutes ago

asked 28 minutes ago

asked 42 minutes ago

asked 43 minutes ago

asked 43 minutes ago