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Nat2105 [25]
2 years ago
13

Calculate the mass of CaCl2•2H2O required to make 100.0 mL of a 0.100 M solution. Each of the calculations below will take you t

hrough the necessary steps. You will be asked to show your answer and calculations for each. Calculate the moles of CaCl2•2H2O in 100.0 mL of a 0.100 M solution Enter your answer:
Chemistry
1 answer:
Zolol [24]2 years ago
4 0

Answer:

The mass is 1.4701 grams and the moles is 0.01.

Explanation:

Based on the given question, the volume of the solution is 100 ml or 0.1 L and the molarity of the solution is 0.100 M. The moles of the solute (in the given case calcium chloride dihydride (CaCl2. H2O) can be determined by using the formula,  

Molarity = moles of solute/volume of solution in liters

Now putting the values we get,  

0.100 = moles of solute/0.1000

Moles of solute = 0.100 * 0.1000

= 0.01 moles

The mass of CaCl2.2H2O can be determined by using the formula,  

Moles = mass/molar mass

The molar mass of CaCl2.2H2O is 147.01 gram per mole. Now putting the values we get,  

0.01 = mass / 147.01

Mass = 147.01 * 0.01

= 1.4701 grams.  

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3 years ago
A buffer solution contains 0.306 M C6H5NH3Br and 0.418 M C6H5NH2 (aniline). Determine the pH change when 0.124 mol HCl is added
Ulleksa [173]

<u>Answer:</u> The pH change of the buffer is 0.30

<u>Explanation:</u>

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

pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})

pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})        .....(1)

We are given:

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pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH_{initial}=14-8.99=5.01

To calculate the molarity, we use the equation:

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

Moles hydrochloric acid solution = 0.124 mol

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M

The chemical reaction for aniline and HCl follows the equation:

                   C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-

<u>Initial:</u>           0.418        0.124           0.306

<u>Final:</u>             0.294          -                0.430

Calculating the pOH by using using equation 1:

pK_b = negative logarithm of base dissociation constant of aniline  = 9.13

[C_6H_5NH_3^+]=0.430M

[C_6H_5NH_2]=0.294M

pOH = ?

Putting values in equation 1, we get:

pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29

To calculate pH of the solution, we use the equation:

pH+pOH=14\\pH_{final}=14-9.29=4.71

Calculating the pH change of the solution:

\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30

Hence, the pH change of the buffer is 0.30

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