Question

Help Plz!!!


22.5cm^3 of sodium hydroxide solution reacted with 25.0cm^3 of 0.100M hydrochloric acid. Calculate the concentration of the sodium hydroxide.

Answer 1

NaOH (aq) + HCl (aq) --> NaCl (aq) + H2O (l)

Answer 2

moles of NaOH : Moles of HCl is 1:1 As u can see the no. of moles HCl is equal to the no. of moles of NaOH.
Using \[C =\frac{ n }{ V}\]
C= concentration
n= no. of moles
V= volume
We can find the no. of moles of HCl right? as we know the concentration and the volume of HCl .
Now subject "n" in that equation
n= CV
Here the concentration is given in moldm^-3
So we have to convert the volume to dm^3 in order to cancel off the units.
\[1cm ^{3}= 10^{-3}dm ^{3}\]
\[V = 25 * 10^{-3} dm ^{3}\]
therefore no. of moles of HCl ;
\[n = 0.1moldm ^{-3} * 25 * 10^{-3} dm ^{3}= 2.5 * 10^{-3} mol\]We know No. of moles of HCl = no. of moles NaOH
So no of moles of NaOH = 2.5 x 10^-3 mol
The information we know about NaOH now is
1)the no. of moles
2) the volume
Here also we need to convert the volume in to cubic decimeter.
so \[V = 22.5 * 10^{-3} dm ^{3}\]
Using the data we know we can find the concentration using the equation
\[C = \frac{ n }{ V }\]
\[C = \frac{ 2.5 * 10^{-3} mol}{ 22.5 * 10^{-3} dm ^{3} } = 0.11M\]

Answer 3

Unless there's a simplification error the answer is correct.

Answer 4

thank you @Rushwr