A 100.0 mL sample of 0.10 M Sr(OH)_2 is titrated with 0.10 M HClO_3. Determine t
ID: 1049647 • Letter: A
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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