How many milliliters of 0.550 m hydriodic acid are needed to react with 5.00 ml of 0.217 m csoh? hi(aq) + csoh(aq) → csi(aq) + h

Question

Elza [17]

3 years ago

13

How many milliliters of 0.550 m hydriodic acid are needed to react with 5.00 ml of 0.217 m csoh? hi(aq) + csoh(aq) → csi(aq) + h

2o(l)?

Chemistry

2 answers:

Olin [163] · Answer 1 · 2022-04-17T14:33:20+03:00

The volume of 0.550 M hydrochloric acid required to react with 5.00 ml of 0.217 M CsOH is $\boxed{19.7{\text{ mL}}}$ .

Further explanation:

Molarity:

The molarity of the solution can be defined as the concentration of the solution and is equal to the number of moles of the solute dissolved in 1 liter of the solution.

The expression of relate molarity (M), volume (V) and the number of moles (n) is as follows:

${\text{M}}=\frac{{{\text{n}}\left({{\text{mol}}}\right)}}{{{\text{V}}\left({\text{L}}\right)}}$ …… (1)

Here, Vis a volume of solution in liters and n is a number of moles of solute.

Rearrange the above equation to calculate the number of moles from molarity.

${\text{n}}\left({{\text{mol}}}\right)=\left({\text{M}}\right)\left({{\text{V}}\left({\text{L}}\right)}\right)$ …… (2)

The given balance reaction is,

${\text{HI}}\left({aq}\right)+{\text{CsOH}}\left({aq}\right)\to{\text{CsI}}\left({aq}\right)+{{\text{H}}_{\text{2}}}{\text{O}}\left(l\right)$

The balanced chemical equation shows that 1 mole of HI reacts with1 mole of CsOH. Therefore, they react in equal amount.

Given molarity of CsOH is 0.217 M, and volume is 5.00 mL.

Substitute these value in equation (2) to calculate the number of moles of CsOH present.

$\begin{aligned}{\text{n}}\left({{\text{mol}}}\right)&=\left({\text{M}}\right)\left({{\text{V}}\left({\text{L}}\right)}\right)\\&=\left({{\text{0}}{\text{.217 }}\frac{{{\text{mol}}}}{{\text{L}}}}\right)\left({5.00{\text{ mL}}\times\frac{{1{\text{ L}}}}{{{{10}^3}{\text{ ml}}}}}\right)\\&=0.0010{\text{ mol}}\\\end{aligned}$

Therefore, number of moles of HCl that is required to react with 0.0010 moles of CsOH is 0.0010 moles of HCl.

To calculate the volume of 0.0010 moles of 0.550 M HCl rearrange the equation (1) as follows:

${\text{V}}\left({\text{L}}\right)=\frac{{{\text{n}}\left({{\text{mol}}}\right)}}{{\text{M}}}$ …… (3)

Substitute the values in equation (3)

$\begin{aligned}{\text{V}}\left( {\text{L}}\right)&=\frac{{{\text{n}}\left({{\text{mol}}}\right)}}{{\text{M}}}\\&=\frac{{0.0010{\text{ mol}}}}{{0.550{\text{ mol/L}}}}\\&=0.00197{\text{ L}}\times\frac{{{{10}^3}{\text{ mL}}}}{{1{\text{ L}}}}\\&=19.7{\text{ mL}}\\\end{aligned}$