Titrations

Question 1

What is titration?

Answer 1

It is a procedure used to determine the concentration of a solution.

Question 2

What is the endpoint of titration?

Answer 2

When the solution changes color.

Question 3

In titration, 27.4 mL of a 0.0154 M solution of Ba(OH)2 is needed to nuetralize 20 mL of HCl. What was the concentration of the acid solution?

Answer 3

0.0211M find the known Moles, balance the equation to determine the molar ratio. Use the information to find the new molarity.

Question 4

Inital solution is 20.00 mL of 0.500 M HCl with 0.500 M NaOH. What is the pH after the addition of 20.20mL of 0.500 M NaOH?

Answer 4

pH = 11.5

1) Identify the equation: HCl + H₂O = H₃0⁺ + Cl^-.
2) identify the number of moles of acid.

Molarty = mol/L , 0.500 M = xmol/ .020 L,

moles of hydronium = .010 mol

3) identify the number of moles of OH in the addition of NaOH,

NaOH + H₂O = H₂O + OH^- + Na⁺

Molarity = mol/ L , 0.050 M = x mol/.0202 L

moles of hydroxide = .0101 mol.

subtract the concentration of moles of acid and the moles of base = 0.001. This is the concentration of OH left over.

3) This is the moles of hydroxide. Now figure out the Molarity so we can caclulate the pH.
0. 2 mL = .0402L = .0024884M .
4) take -log( .002488) = 2.5 = pOH.
5) 14= pH + pOH, ph= 11.5

Question 5

what is the equivalence point in titration?

Answer 5

It is the point where all of the base and acid are neutralized and pH of 7.

Question 6

Henderson hasslebach equation

Answer 6

pH = pKa + log( A^-)/(HA)

Good for buffer solutions.

Question 7

Titration curve of 40.0 mL of 0.100 M of NH₃ with 0.100 M HCl.

A) what is the pH before any acid is added? (Kb_amonium + 1.85 x 10^-5)

B) What is the pH after the addition of 20.0 mL of 0.100 M HCl?

Answer 7

A = pH = 11.11

B =

A) work:

1) write equation:

 NH<sub>3 </sub>+ H<sub>2</sub>O NH<sub>4</sub>  OH<sup>-</sup>

I 0.100 M 0 0

C -x + x +x

E 0.10 -x x x

2) K_b= (NH₄₎)(OH) / NH₃ => 1.8x10^-5 = x²/ 0.100 = 1.3 x10^-3
3) pOH = -log( 1.3x10^-3) = 2.89
4) pOH + pH = 14, pH = 11.11

Part B work:

1) write the equation

NH₃+ HCl —> NH₄⁺ + Cl^-

2) Figure out the mole ratio

Molarity = moles/L moles of NH₃ = 0.004 mol

moles of HCl = .002

3) acids/baes neutralize each other on a one to one basis. Subtract the moles of acid from the moles of base. Take the absolute value and figure out molarity.

.004-.002 = 0.002 moles of NH₃

M of NH₃ = 0.002 moles/ 0.06L = .033M = concentration of NH₃

4) do the weak base calculations:

NH3 + H2O < —–> NH4 OH-

I .033 0 0

C -x +x +x

E 0.033-x

Kb = (B)(H)/ BH = 1.8x10^-5 = x²/ .033 = 6.0 x 10^-7 = concentration of base

4) Find pH, -log(Molarity)
- log(6.0x10^-7) = pOH = 7

7 + pH = 14.