Q.10.108 - An Introduction to General, Organic, and Biological Chemistry

Question

Determine each of the following for a 0.100 M HBr solution: 

a. $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

b. pH

c. the balanced chemical equation for the reaction with LiOH

d. millilitres of the HBr solution required to neutralize 36.0 mL of a 0.250 mL of LiOH solution

Step-by-Step Solution

Answer

Part (a). $0.10 M$

Part (b) . $1$

Part (c). $\mathrm{HBr} \left(aq\right)+\mathrm{LiOH} \left(aq\right)\to \mathrm{LiBr} \left(aq\right)+{\mathrm{H}}_2\mathrm{O} \left(l\right)$

Part (d). $90 ml$

Step 1: Given Information (Part a)

A chemical reaction is an interaction wherein at least one substance, likewise called reactants, is switched over completely to at least one distinct substance, known as items.

Step 2: Explanation Part (a)

Considering the reaction

$\mathrm{HBr} \left(aq\right)+{\mathrm{H}}_2\mathrm{O} \left(l\right)\to {\mathrm{Br}}^{-}\left(aq\right)+{\mathrm{H}}_3{\mathrm{O}}^{+}\left(aq\right)$

From the reaction, we get

$\left[\mathrm{HBr}\right]=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=0.100 \mathrm{M}$as HBr completely ionizes.

Step 3: Explanation Part (b)

Using,

$\mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

$$\begin{gathered} \mathrm{pH}=-\log \left[0.100\right] \\ =1 \end{gathered}$$

Step 4: Explanation Part (c)

The chemical reaction is - $\mathrm{HBr} \left(aq\right)+\mathrm{LiOH} \left(aq\right)\to \mathrm{LiBr} \left(aq\right)+{\mathrm{H}}_2\mathrm{O} \left(l\right)$

Step 5: Explanation Part (d)

Given,

Volume is $V_1$=$36 ml$of $M_2$=$0.250 \mathrm{M} \mathrm{LiOH}{\text{ and M}}_1\text{=}0.100 \mathrm{M} \mathrm{HBr}$ solution.

Using,

$M_1{V}_1=M_2{V}_2$

${\mathrm{V}}_2=\frac{0.250 \mathrm{M}\times 36 \mathrm{mL}}{01 \mathrm{M}}=90 mL$

Hence, $90$ millilitres of the HBr solution are required.