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rjkz [21]
3 years ago
15

The absorbance of a garbanzo bean solution that had been diluted by a factor of three was 0.528. what was the concentration of t

he iron in the original solution (before dilution)?
Chemistry
1 answer:
Serjik [45]3 years ago
6 0
The Beer-Lambert law states that A = E*c*l where A is absorbance, E is the molar absorbance coeffecient, c is concentration and l is path length. Therefore the absorbance is directly proportional to concentration, and by increasing the concentration by a factor of 3, absorbance will increase by a factor of 3 giving   A = 1.584
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6 0
3 years ago
Calculate the pH of adding 20 mL of 1 M NaOH solution to 100 mL of a 1 M acetic acid (CH3COOH) solution and 880 mL of distilled
Dafna1 [17]

<u>Answer:</u> The pH of the solution is 4.14

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

  • <u>For NaOH:</u>

Molarity of NaOH = 1 M

Volume of solution = 20 mL

Putting values in above equation, we get:

1M=\frac{\text{Moles of NaOH}\times 1000}{20mL}\\\\\text{Moles of NaOH}=\frac{1\times 20}{1000}=0.02mol

  • <u>For acetic acid:</u>

Molarity of acetic acid solution = 0.26 M

Volume of solution = 100 mL

Putting values in above equation, we get:

1M=\frac{\text{Moles of acetic acid}\times 1000}{100mL}\\\\\text{Moles of acetic acid}=\frac{1\times 100}{1000}=0.1mol

The chemical reaction for NaOH and acetic acid follows the equation:

                   CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O

<u>Initial:</u>           0.100            0.020

<u>Final:</u>            0.080               -              0.020            

Volume of solution = 20 + 100 = 120 mL = 0.120 L    (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[salt]}{[acid]})

pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})

We are given:

pK_a = negative logarithm of acid dissociation constant of acetic acid = 4.74

[CH_3COONa]=\frac{0.020}{0.120}

[CH_3COOH]=\frac{0.080}{0.120}

pH = ?

Putting values in above equation, we get:

pH=4.74+\log(\frac{0.020/0.120}{0.080/0.120})\\\\pH=4.14

Hence, the pH of the solution is 4.14

5 0
3 years ago
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