Question

How many mL of 2.5M HCl would be needed to completely neutralize a standard solution of 0.53M NaOH in a titration

Answer 1

Amount of HCL = 0.00318 L of 3.18 ml

Step-by-step explanation:

Given:

HCL = 2.5 M

NaOH = 0.53 M

Amount of NaOH = 15 ml = 0.015 L

Find:

Amount of HCL

Computation:

HCL react with NaOH

HCl + NaOH ⇒ NaCl + H₂O

So,

Number of moles = Molarity × volume

Number of moles of NaOH = 0.53 × 0.015

Number of moles of NaOH = 0.00795 moles

So,

Number of moles of HCl needed = 0.00795 mol es

So,

Volume = No. of moles / Molarity

Amount of HCL = 0.00795 / 2.5

Amount of HCL = 0.00318 L of 3.18 ml