Question 7.8: Solutions of NaOH and HCl react according to the following e...

a. The known quantity is 0.110 mol of NaOH, and the unit of the unknown quantity is a volume which we will express as L NaOH soln.
Step 1. 0.110 mol NaOH

Step 2. 0.110 mol NaOH                                               = L  NaOH soln.

Step 3. 0.110    \cancel{mol}    \cancel{NaOH} × \frac{1  L  NaOH  soln.}{0.250  \cancel{mol}    \cancel{NaOH} } =  L  NaOH  soln.

The factor \frac{1  L  NaOH  soln.}{0.250  mol    NaOH } came from the known molarity of the NaOH solution as  described above.

Step 4. 0.110 × \frac{1  L  NaOH  soln.}{0.250 } = 0.44  L  NaOH  soln. =  0.440  L  NaOH  soln.

The calculator answer of 0.44 was rounded by adding a trailing zero so the answer had three significant figures to match the three significant figures in 0.110 and 0.250. The number 1 is an exact counting number.

b. The known quantity is 0.150 mol HCl, and the unit of the unknown quantity is the volume of NaOH solution which we will express as L NaOH soln.

Step 1. 0.150 mol HCl
Step 2. 0.150 mol HCl                                                              = L NaOH soln.

Step 3.  0.150   \cancel{mol}  \cancel{HCl}  × \frac{ 1  \cancel{mol}    \cancel{NaOH}}{1   \cancel{mol}  \cancel{HCl}} × \frac{1  L  NaOH  soln.}{0.200  \cancel{mol}  \cancel{HCl}}  =L NaOH soln.

Note that two factors were needed to cancel the unit of the known quantity and generate the unit of the unknown quantity. The first factor came from statement 2 for the reaction:

NaOH(aq) + HCl(aq) → NaCl(aq) + H_{2}O(ℓ)
Statement 2:   1  mol  NaOH + 1  mol  HCl → 1  mol  NaCl + 1  mole  H_{2}O

The second factor came from the 0.200 M concentration of the NaOH solution. Two factors were possible: \frac{0.200  mol   NaOH}{1  L  NaOH  soln.}  and  \frac{1  L  NaOH  soln.}{0.200  mol  NaOH}. The second factor was used because it cancelled the mol NaOH unit and generated the L NaOH soln. unit.

Step 4. 0.150 × \frac{1}{1} × \frac{1  L  NaOH  soln.}{0.200  } = 0.75  L  NaOH  soln. = 0.750  L  NaOH  soln.
Once again, a zero was added to the calculator answer of 0.75 to give an answer with three significant figures to match the three in 0.150 and 0.200. Both of the 1 numbers are exact counting numbers.

c. The known quantity is 25.0 mL of a 0.225 M HCl soln., and the unit of the unknown is the volume of a 0.185 M NaOH soln. which we will express as mL NaOH soln. to match the mL unit of the known quantity.
Step 1. 25.0 mL HCl soln.
Step 2. 25.0 mL HCl soln.            = mL NaOH soln.

Step 3. 25.0   \cancel{mL}   \cancel{HCl}   \cancel{soln}. × \frac{0.225  \cancel{ mol}  \cancel{ HCl}}{1000  \cancel{mL}   \cancel{HCl}   \cancel{soln}.}  \\ × \frac{ 1  \cancel{mol}    \cancel{NaOH}}{1   \cancel{mol}  \cancel{HCl}} × \frac{1000  mL  NaOH  soln.}{0.185  \cancel{mol}    \cancel{NaOH}} = mL  NaOH soln.

Note that three factors were needed. The first came from the molarity of the HCl solution with the solution volume expressed as 1000 mL rather than 1 L. The second factor came from statement 2 for the reaction as was done in part b. The third factor came from the molarity of the NaOH solution with the solution volume expressed as 1000 mL rather than 1 L.

Step 4.  25.0 × \frac{0.225}{1000}× \frac{1}{1}×\frac{1000  mL  NaOH  soln.}{0.185} = 30.4  mL  NaOH  soln.