Consider the reaction represented by the question below: KOH_(aq) + HCI_(aq) → KCI_(aq) + H₂O_(l) What volume of 0.25 mol dm^-3 KOH would be required to completely neutralize 40 cm³ of 0.10 mol dm ^-3 HCI?

Samuel.og

4 years ago

CaVa=CbVb
Ca=0.10mol/dm3
Va=40cm3
Cb=0.25mol/dm3
Vb=?
Vb=CaVa/Cb
=0.10*40/0.25=16

princessnwachukwu

6 years ago

please can the admins of myschool solve the questions so students can get their explanations

Tolujedjames

6 years ago

Can the solving for thi please be shown?

BrodaChoco

1 year ago

molarity = conc. × volume
For the acid,
Convert 40 cm³ to dm 40 × (0.1dm)³ = 40 × 10-³dm³
Molarity = 0.10moldm-³ × 40 × 10-³dm³
= 4 × 10-³mol
Since from the equation, the molarity is 1:1, it therefore means 4 × 10-³mol is required of the base
Using, Molarity = conc. × volume
Volume = conc./ Molarity
V = 4 × 10-³mol/0.25moldm-³
= 16dm³
From the options above it's obvious there was no conversion from cm³ to dm³ so I can easily pick 16cm³

Temiolap

7 years ago

Pls guide me oh