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Fynjy0 [20]
3 years ago
14

Acetanilide has a solubility of 1.00 g/185 ml in 25 oc water and 1.00 g/20 ml in 100 oc water. what is the minimum volume of wat

er needed to recrystallize 500 mg of acetanilide? how much acetanilide will remain dissolved after crystallization at 25 oc?
Chemistry
1 answer:
Likurg_2 [28]3 years ago
6 0

In the crystallization process the solid compound is dissolved in the solvent at elevated temperature and the crystallize product obtained by slow cooling of the  solution. Here the solubility of acetanilide at 100°C is 1g per 20mL of water. Thus to dissolve 500mg of acetanilide at high temperature that is 100°C we need 10mL of water.

Now at 25°C after the re-crystallization there will be some amount of dissolve acetanilide. Which can be calculated as- 185mL of water is needed to dissolve 1g or 1000mg of acetanilide at 25°C. Thus in 10mL of water there will be \frac{10X1000}{185}=54.05gmg of acetanilide.    

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2 years ago
You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.8
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Complete Question

You determine that it takes 26.0 mL of base to neutralize a sample of your unknown acid solution. The pH of the solution was 7.82 when exactly 13 mL of base had been added, you notice that the concentration of the unknown acid was 0.1 M. What is the pKa of your unknown acid?

Answer:

The pK_a value is pK_a  =7.82

Explanation:

From the question we are told

    The volume of base is  V_B = 26.mL = 0.0260L

     The pH of solution is  pH =  7.82

      The concentration of the acid is C_A = 0.1M

From the pH we can see that the titration is between a strong base and a weak acid

 Let assume that the the volume of acid is  V_A = 18mL= 0.018L

Generally the concentration of base

                    C_B = \frac{C_AV_A}{C_B}

Substituting value  

                     C_B = \frac{0.1 * 0.01800}{0.0260}

                    C_B= 0.0692M

When 13mL of the base is added a buffer is formed

The chemical equation of the reaction is

           HA_{(aq)} + OH^-_{(aq)} --------> A^{+}_{(aq)} + H_2 O_{(l)}

Now before the reaction the number of mole of base is  

            No \ of \ moles[N_B]  =  C_B * V_B

Substituting value  

                    N_B = 0.01300 * 0.0692

                         = 0.0009 \ moles    

                                 

Now before the reaction the number of mole of acid is  

            No \ of \ moles  =  C_B * V_B

Substituting value  

                    N_A = 0.01800 *0.1

                         = 0.001800 \ moles

Now after the reaction the number of moles of  base is  zero  i.e has been used up

    this mathematically represented as

                         N_B ' = N_B - N_B = 0

    The  number of moles of acid is  

             N_A ' = N_A  - N_B

                   = 0.0009\ moles

The pH of this reaction can be mathematically represented as

                 pH  = pK_a + log \frac{[base]}{[acid]}

Substituting values

                  7.82 = pK_a +log \frac{0.0009}{0.0009}

                  pK_a  =7.82        

                     

             

                                 

       

           

                     

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<span>Physical change, heat caused this.

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</span>
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3 years ago
Read 2 more answers
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