A 100.0 mL sample of 0.10 M Sr(OH)_2 is titrated with 0.10 M HClO_3. Determine t

Question

A 100.0 mL sample of 0.10 M Sr(OH)_2 is titrated with 0.10 M HClO_3. Determine the pH of the solution before the addition of any HClO_3. Determine the pH of the solution after the addition of 100.0 mL HClO_2. Determine the volume of HClO_3 required to reach the equivalence point Determine the pH of the solution after the addition of 300.0 mL HClO_3. Sketch and label (axes, equivalence point) the titration curve. Phenoiphthalein is used as an indicator. Label the end point and expected color change.

Explanation / Answer

Sr(OH)2 is a strong base,

It dissociation is as follows:

Sr(OH)2 ----------> Sr2+ +2OH- all are in aqueous state.

Calculate the [OH- ] as follows:

[OH- ]=2*0.1 M=0.2 M

-log[OH- ]=pOH

pOH= -log[0.2]=0.6989

pH+pOH=14

pH=14-0.6989=13.30

Therefor, the initial pH of the solution is 13.30.

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Calculate the moles of OH- that remains after the addition of 100 mL of HClO3 is added:

moles of H+ = 0.1 mol/L x 0.1 L = 0.01 mol

moles of [OH-]=0.2 mol/L x 0.1 L = 0.02 mol

Remaining OH- =0.02 mol - 0.01 mol = 0.01 mol

3) Calculate the [OH-]:

The totl volume is 200 mL=0.2 L

[OH-] = 0.01 mol / 0.2 L = 0.05 M

pOH = -log 0.05= 1.30

pH = 14 - 1.30 = 12.69

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Chemical equation:

2HClO3 + Sr(OH)2 --->  Sr(ClO3)2 + 2 H2O

moles of Sr(OH)2:

Moles of  Sr(OH)2=0.1 mol/L*0.1 L= 0.01 moles

The molar ratio between HClO3 and Sr(OH)2 is 2:1

Two moles of HClO3 are used up for every one mole of Sr(OH)2 reacted.

Moles of HClO3 used:2*0.01 moles=0.02 moles

Volume of HCl required:

moles = MV

0.02 moles = (0.1 mol/L) (x)

x = 0.2 L

x = 200 mL

Therefore, the volume of acid required at the equivalence point is 200 mL.

Calculate the moles of OH- that remains after the addition of 200 mL of HClO3 is added:

moles of H+ = 0.1 mol/L x 0.2 L = 0.02 mol

moles of [OH-]=0.2 mol/L x 0.1 L = 0.02 mol

Remaining OH- = 0.02 mol - 0.02 mol =0 mol

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Calculate the moles of OH- that remains after the addition of 300 mL of HClO3 is added:

moles of H+ = 0.1 mol/L x 0.3 L = 0.03 mol

moles of [OH-]=0.2 mol/L x 0.1 L = 0.02 mol

Remaining OH- = 0.02 mol - 0.01 mol = 0.01 mol

3) Calculate the [OH-]:

The total volume is 400 mL=0.4 L

[OH-] = 0.01 mol / 0.4 L = 0.025 M

pOH = -log 0.025= 2.602

pH = 14 - 2.602 = 11.39

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The color change is reddish pink

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