Chapter U4.87, Problem 7E

(a)

Interpretation Introduction

Interpretation:

Number of moles of H+ in 0.75 L of a 0.10 M HCl solution must be calculated.

Concept Introduction :

Molarity of HCl solution is 0.10 means there is 0.10 mol of HCl in 1L solution.

Answer to Problem 7E

0.075 mole of H+ are there in 0.75 L of a 0.10 M HCl solution.

Explanation of Solution

From molarity, moles can be calculated as follows:

  $n=M\times V$

Here, M and V are molarity and volume of the solution in L.

So, moles of $\text{HCl=0}\text{.10×0}\text{.75=0}\text{.075}\text{.}$

As HCl is a strong acid it will produce quantitative H⁺. So moles of ${\text{H}}^{\text{+}}\text{=moles of HCl=0}\text{.075}\text{.}$

  ${\text{H}}^{\text{+}}\text{=moles of HCl=0}\text{.075}\text{.}$

(b)

Interpretation Introduction

Interpretation:

If 0.35 L water is added then the new concentration of HCl must be calculated.

Concept Introduction :

If water is added to an acid solution, the concentration decreases due to increase in volume of the solution keeping the amount of solute same.

Answer to Problem 7E

Concentration of HCl will be 0.068 M.

Explanation of Solution

The relation between initial and final strength and volume will be:

  ${\text{V}}_{\text{1}}{\text{S}}_{\text{1}}{\text{=V}}_{\text{2}}{\text{S}}_{\text{2}}$

Here V₁, S₁, V₂ and S₂ are initial volume, initial strength, final volume and final strength for the HCl solution respectively.

Putting the values it can be written as

  $0.75\times 0.10=1.1\times S_2$

Or,

  ${\text{S}}_{\text{2}}\text{=}\frac{\text{0}\text{.075}}{\text{1}\text{.1}}\text{=0}\text{.068}\text{.}$

So, final concentration of HCl is 0.068 M.

(c)

Interpretation Introduction

Interpretation:

The pH of the diluted HCl solution needs to be calculated.

Concept Introduction :

The pH of the solution can be calculated as follows:

  $\text{pH=-log}\left[{\text{H}}^{\text{+}}\right]\text{.}$

With dilution of acid solution, pH increases.

Answer to Problem 7E

pH will be 1.17.

Explanation of Solution

New concentration of HCl solution is 0.068 M after 0.35 L water is added.

Now,

  $\left[{\text{H}}^{\text{+}}\right]\text{=}\left[\text{HCl}\right]\text{=0}\text{.068 M}$

Thus,

  $\text{pH=-log}\left[{\text{H}}^{\text{+}}\right]\text{=-log}\left[\text{0}\text{.068}\right]\text{=1}\text{.17}\text{.}$