Use the concentrations and volumes of hydrogen peroxide and KI before mixing (Table 1),
calculate the post-mixing concentrations.
Before mixing:
[H2O2] = 0.88M
[KI] = 0.500M
Table 1: before mixing
1. Vol H2O2 = 4.0 mL, [H2O2] = 0.88M
2. Vol KI = 1.0 mL, [KI] = 0.500 M
units: M
I have kept trying to do this one but I am not getting the right answer. the answer is given (after mixing), [H2O2]=0.70M, [KI]=0.100M
i donot know this one next
[H2O2] before = 0.88M
volume H2O2 before mix is 4 mL
total volume after mix is 5 (4 from H2O2 + 1 from KI) so the H2O2 has been diluted from 4 to 5 mL.'
0.88 x 4/5 = ?
KI before mix is 0.500.
volume KI before mix is 1 mL.
total volume after mix is 5 (1 from KI + 4 from H2O2). The 0.500 M KI has been diluted from 1 mL to 5 mL so
final [KI] is 0.500 x 1/5 = ?
To calculate the post-mixing concentrations of hydrogen peroxide (H2O2) and potassium iodide (KI), we need to use the concept of dilution.
Dilution formula:
(C1)(V1) = (C2)(V2)
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
Let's first calculate the final volume after mixing:
Final volume = Volume H2O2 + Volume KI
= 4.0 mL + 1.0 mL
= 5.0 mL
Now, let's calculate the final concentration of H2O2:
(C1)(V1) = (C2)(V2)
(0.88M)(4.0 mL) = (C2)(5.0 mL)
3.52 = 5C2
C2 = 3.52/5
C2 = 0.704 M
Therefore, the post-mixing concentration of H2O2 is 0.704 M or approximately 0.70 M.
Now, let's calculate the final concentration of KI:
(C1)(V1) = (C2)(V2)
(0.500M)(1.0 mL) = (C2)(5.0 mL)
0.500 = 5C2
C2 = 0.500/5
C2 = 0.100 M
Therefore, the post-mixing concentration of KI is 0.100 M.
So, the final concentrations after mixing are:
[H2O2] = 0.70 M
[KI] = 0.100 M
To calculate the post-mixing concentrations of hydrogen peroxide ([H2O2]) and potassium iodide ([KI]), we need to consider the initial concentrations and volumes of each reactant.
Let's calculate the moles of each compound before mixing:
For hydrogen peroxide:
moles of H2O2 = concentration of H2O2 × volume of H2O2
= 0.88 M × 4.0 mL
For potassium iodide:
moles of KI = concentration of KI × volume of KI
= 0.500 M × 1.0 mL
Since the reaction between hydrogen peroxide and potassium iodide is a redox reaction, it is balanced by a 2:1 stoichiometry:
H2O2 + 2KI → 2KOH + I2 + H2O
From the balanced equation, we see that for every 2 moles of KI, 1 mole of H2O2 reacts. Therefore, the moles of H2O2 and KI will decrease by the same factor upon reaction.
Now let's calculate the post-mixing concentrations:
New moles of H2O2 = initial moles of H2O2 - (0.5 × initial moles of KI)
= (0.88 M × 4.0 mL) - (0.5 × (0.5 M × 1.0 mL))
New concentration of H2O2 = new moles of H2O2 / (volume of H2O2 + volume of KI)
= new moles of H2O2 / (4.0 mL + 1.0 mL)
New moles of KI = initial moles of KI - (2 × initial moles of H2O2)
= (0.5 M × 1.0 mL) - (2 × (0.88 M × 4.0 mL))
New concentration of KI = new moles of KI / (volume of H2O2 + volume of KI)
= new moles of KI / (4.0 mL + 1.0 mL)
By substituting in the appropriate values and calculating, we find:
New concentration of H2O2 = 0.70 M
New concentration of KI = 0.100 M
Therefore, after mixing, the concentrations of hydrogen peroxide and potassium iodide are [H2O2] = 0.70 M and [KI] = 0.100 M, respectively.