0.6 Neutralization reaction  (Page 2/2)

Problem : A 25 ml of nitric acid taken from a stock volume of 1 litre neutralizes 50 ml of 0 .1N NaOH solution. Determine the mass of nitric acid in the stock volume.

Solution : Let us denote nitric acid and sodium hydroxide by subscripts "1" and "2" respectively. Applying neutralization equation,

$$N_1{V}_1=N_2{V}_2$$

$$\Rightarrow N_{1}X25=0.1X50$$

$$\Rightarrow N_1=\frac{5}{25}=0.2$$

The normality of sample and the stock volume is same. Hence, normality of 1 litre stock volume is 0.2N. Using formula,

$$\Rightarrow \text{geq}=NV=0.2X1=0.2=\frac{g_B}{E}=\frac{x{g}_B}{M_O}$$

Valence factor of nitric acid is 1 as it has one furnishable hydrogen. Therefore,

$$\Rightarrow g_B=\frac{0.2M_O}{x}=\frac{0.2X\left(1+14+3X16\right)}{1}=12.6gm$$

Problem : 18 ml of 0.1N $H_{2}S{O}_4$ is neutralized by 20 ml of a NaOH solution. On the other hand, 10 ml of oxalic acid is required to neutralize the same volume of NaOH solution. Determine the mass of oxalic acid crystals $\left\{{\left(COOH\right)}_2.2H_{2}O\right\}$ used.

Solution : A solution of known concentration of sulphuric acid neutralizes sodium hydroxide solution of unknown concentration. Applying neutralization equation, we determine normality of sodium hydroxide solution :

$$N_1{V}_1=N_2{V}_2$$

$$\Rightarrow 0.1X18=N_{2}X20$$

$$\Rightarrow N_2=\frac{1.8}{20}=0.09N$$

Now, a solution of known concentration of sodium hydroxide neutralizes oxalic acid solution of unknown concentration. Again applying neutralization equation, we determine normality of oxalic acid solution :

$$N_1{V}_1=N_2{V}_2$$

$$\Rightarrow 0.09X20=N_{2}X10$$

$$\Rightarrow N_2=\frac{1.8}{10}=0.18N$$

Using formula,

$$\Rightarrow \text{geq}=NV=\frac{0.19X10}{1000}=\frac{1.9}{1000}=0.0019=\frac{g_B}{E}=\frac{x{g}_B}{M_O}$$

Valence factor of oxalic acid is 2 as it has two furnishable hydrogens. Therefore,

$$\Rightarrow g_B=\frac{0.0019M_O}{x}=\frac{0.0019X\{2X\left(12+2X16+1\right)+2X18\}}{2}=0.1197gm$$

Problem : 5 ml each of 2N hydrochloric and 3N nitric acids volumes are mixed with a certain volume of 5N sulphuric and the resulting solution is made up to 1 litre. A volume of 25 ml of this solution neutralizes 50 ml of sodium carbonate solution containing 1 gm of $N{a}_{2}C{O}_3.10H_{2}O$ in water. Determine the volume of sulphuric acid in the mixture.

Solution : A combination of different acids are used here. The mixture is then diluted up to 1 litre. Let the volume of sulphuric acid used is x ml. Here, milli-gram equivalents of the mixture is :

$$\Rightarrow \text{meq}=5X2+5X3+5x=25+5x$$

Applying equation of dilution,

$$N_1{V}_1=N_2{V}_2$$

$$\Rightarrow 25+5x=N_{2}X1000$$

$$\Rightarrow N_2=\frac{25+5x}{1000}$$

We see here that normality of acid solution has one unknown. 25 ml of diluted acid neutralizes 50 ml of sodium carbonate solution. Clearly, we need to calculate normality of sodium carbonate used. We see here that strength of sodium carbonate solution is given (1 gm/50 ml = 20 gm/l). Using relation,

$$S=NE$$

The valence factor of sodium carbonate is 2 as its cation or anion has 2 electronic charge.

$$\Rightarrow N=\frac{S}{E}=\frac{xS}{M_O}=\frac{2X20}{\left(2X23+12+3X16+10X18\right)}=\frac{40}{286}=0.14N$$

Now, using neutralization equation :

$$N_1{V}_1=N_2{V}_2$$

$$\Rightarrow \frac{25+5x}{1000}X25=0.14X50$$

$$\Rightarrow 25+5x=0.28X1000=280$$

$$\Rightarrow 5x=280-25=255$$

$$\Rightarrow x=\frac{255}{5}=51ml$$

Problem : 1 gm of oleum is diluted in water. The resulting solution requires 45 ml of 0.5N sodium hydroxide solution for neutralization. Determine the mass of free $S{O}_3$ in the oleum.

Solution : Let mass of $S{O}_3$ in oleum is x. Then, mass of sulphuric acid in the oleum is 1-x. We can find gram-equivalents of the sulphuric acid and oleum provided we know valence factors. The valence factor of each of them is 2. Hence, equivalent weights of sulphuric acid and Sulphur trioxide are 98/2=49 and 80/2=40 respectively. Applying neutralization equation,

$$\text{geq of}{H}_{2}S{O}_4+\text{geq of}S{O}_3=\text{geq of}base$$

$$\frac{1-x}{49}+\frac{x}{40}=NV=\frac{0.5X45}{1000}=\frac{22.5}{1000}$$

$$\Rightarrow 40-40x+49x=49X40X\frac{22.5}{1000}=44.100$$

$$\Rightarrow 9x=4.41$$

$$\Rightarrow x=\frac{4.41}{9}=0.5gm$$