Question

How many millilitres of 0.100 mo(l)/(L) HCl are required to neutralize 25.0 mL

of 0.100 mo(l)/(L) KOH? (25 mL)

Answer 1

$n = C×V$

$n = 0.1M × 0.025L$

n = 0.0025 mol KOH

HCl+КOH=NaCl+H₂O

the coefficients are the same so the amount of the substance is the same

n(HCl)=0.0025

$V = \frac{n}{C}$

$V = 0.0025 mol/0.1M$

V = 0.025 L = 25 mL HCl