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

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`

Step-by-step 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 = (10)/(1000) = 0.010 L`

The volume of NaOH is
`V_(NaOH_1 ) = 30 cm^3 = (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 = (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 = (Q)/(c_p * \Delta T)`

substituting values

`m = (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)} = (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)} = (0.9)/(36.46 * 0.01)}`

`M__(HCl)} =2.468 M`