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

Short answer

Part (a). $0.10M$

Part (b) . $1$

Part (c). localid="1654602027864" $\mathrm{HBr}\left(aq\right)+\mathrm{LiOH}\left(aq\right)\to \mathrm{LiBr}\left(aq\right)+{\mathrm{H}}_2\mathrm{O}\left(l\right)$

Part (d). localid="1654602034131" $90ml$

Step-by-step solution

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.

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.

Explanation Part (b)

Using,

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

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

Explanation Part (c)

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

Explanation Part (d)

Given,

Volume is $V_1$=localid="1654602108020" $36ml$of $M_2$=localid="1654602118429" $0.250\mathrm{M}\mathrm{LiOH}{\text{andM}}_1\text{=}0.100\mathrm{M}\mathrm{HBr}$solution.

Using,

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

localid="1654602127161" ${\mathrm{V}}_2=\frac{0.250\mathrm{M}\times 36\mathrm{mL}}{01\mathrm{M}}=90mL$

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