star_girl one year ago 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.

• This Question is Open
1. star_girl

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

2. Rushwr

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 $C = \frac{ n }{ V }$ $C = \frac{ 2.5 * 10^{-3} mol}{ 22.5 * 10^{-3} dm ^{3} } = 0.11M$

3. Rushwr

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

4. star_girl

thank you @Rushwr