Question

Question 1

If we mixed 45cm^3 of 2m NaOH with 20cm^3 of 1m HCL and obtained a temperature difference of 9.4°C, what would be the heat of reaction, the enthalpy?.

Question 2

If we added 10cm^3 HCL of an unknown concentration to 30cm^3 of 2M NaOH and obtained a temperature change of 4.5°C and then added 20cm^3 of the same unknown acid to 30cm^3 of 2M NaOH and obtained a temperature change of 9.0°C, what is the concentration of the acid? Use the enthalpy value from question 1

Answer 1

Answer:

1

The heat of reaction is $Q = 170.92 \ J$

The enthalpy is $\Delta H = -170.91 \ J$

2

The concentration of HCl is  $M__{HCl}} =2.468 M$

Explanation:

From the question we are told that

      The volume of  $NaOH$ is  $V_{NaOH} = 45 \ cm^3 = \frac{45} {1000} = 0.045 L$

       The number of concentration of  $NaOH$ is  $M__{NaOH}} = 2M$

       The volume of  HCl is $V_{HCl} = 20 cm^3$

        The number of concentration of  $HCl$ is  $M__{HCl}} = 1 M$

        The temperature difference is $\Delta T = 9.4 ^o C$

Now the heat of reaction is mathematically represented as

         $Q = m * c_p * \Delta T$

Where

      $c_p$ is the specific heat of water with value  $c_p = 4200 J /kg ^o C$

      $m = m__{NaOH}} + m__{HCl}}$

Now   $m__{NaOH}} = V_{NaOH} * M_{NaOH} * Z_{NaOH}$

where  $Z_{NaOH}$ is the molar mass of NaOH  with the value of  0.04 kg/mol

    So    $m__{NaOH}} = 0.045 * 2 * 0.040$

            $m__{NaOH}} =0.0036\ kg$  

While  

        $m__{HCl}} = V_{HCl} * M_{HCl} + Z_{HCl}$

Where $Z_{HCl}$ is the molar mass of $HCl$ with the value of  0.03646 kg/mol

        $m__{NaOH}} = 0.020 * 1 * 0.03646$

       $m__{NaOH}} = 0.000729 kg$

So

     $m = 0.0036 + 0.000729$

     $m = 0.00433$

=>  $Q = 0.00433 * 4200 * 9.4$

     $Q = 170.92 \ J$

The enthalpy is mathematically represented as

       $\Delta H = - Q$

=>    $\Delta H = -170.91 \ J$

From the second question we are told that

      The volume of  HCl is $V_{HCl_1} = 10cm^3 = \frac{10}{1000} = 0.010 L$

       The volume of  NaOH  is  $V_{NaOH_1 } = 30 cm^3 = \frac{30}{1000} = 0.03 L$

      The concentration of NaOH is  $M_{NaOH} = 2 M$

       The first temperature change is  $\Delta T = 4.5 ^oC$

       The second volume of  $V_{HCl_2} = 20 cm^3 = \frac{20}{1000 } = 0.020 m^3$

The mass of NaOH is

       $m__{NaOH}} = V_{NaOH} * M_{NaOH} * Z_{NaOH}$

substituting values

       $m__{NaOH}} = 0.03 * 2 * 40$

       $m__{NaOH}} = 3.6 \ g$

The mass of the product formed is

       $m = \frac{Q}{c_p * \Delta T}$

substituting values  

     $m = \frac{170.91}{4200 * 9} * 1000$    

The multiplication by 1000 is to convert it from kg to grams

      $m = 4.5 g$

Now the mass of HCl is  

      $m__{HCl}} = m - m__{NaOh}}$

substituting values    

        $m__{HCl}} = 4.5 -3.6$

        $m__{HCl}} = 0.9 \ g$

Now the concentration of HCl is  

         $M__{HCl}} = \frac{m_{HCl}}{(Z_{HCl} * *1000) * V_{HCl_1}}}$

The multiplication of $Z_{HCl}$ is to convert it from kg/mol to g/mol

          $M__{HCl}} = \frac{0.9}{36.46 * 0.01}}$

          $M__{HCl}} =2.468 M$