Question

Determine the volume of 0.150 M NaOH solution required to neutralize each sample of hydrochloric acid. The neutralization reaction is: NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq) 25 mL of a 0.150 M HCl solution 55 mL of a 0.055 M HCl solution 175 mL of a 0.885 M HCl solution

Answer 1

The volume 0.150+175+0.885+0.055+25!=201.09

Answer 2

Answer: The Volume of NaOH required to neutralize

1) 25 mL of a 0.150M HCl solution is 25 mL

2) 55 mL of a 0.055M HCl solution is 20.2 mL

3) 175 mL of 0.885M HCl solution is 1032.5 mL

Explanation: Neutralization reactions are defined as the reactions when an acid reacts with a base to form a salt and water.

For the reaction of HCl and NaOH,

$NaOH(aq.)+HCl(aq.)\rightarrow NaCl(aq.)+H_2O(l)$

To calculate the molarity of NaOH, we use the formula:

$M_{NaOH}\times V_{NaOH}=M_{HCl}\times V_{HCl}$

1.) 25 mL of a 0.150M of HCl solution.

$M_{NaOH}=0.150M;V_{HCl}=25mL;M_{HCl}=0.150M$

Putting the values in above equation, we get

$0.150\times V_{NaOH}=0.150\times 25\\V_{NaOH}=25mL$

2.) 55 mL of a 0.055M of HCl solution.

$M_{NaOH}=0.150M;V_{HCl}=55mL;M_{HCl}=0.055M$

Putting the values in above equation, we get

$0.150\times V_{NaOH}=0.055\times 55\\V_{NaOH}=20.2mL$

3.) 175 mL of a 0.885M of HCl solution.

$M_{NaOH}=0.150M;V_{HCl}=175mL;M_{HCl}=0.885M$

Putting the values in above equation, we get

$0.150\times V_{NaOH}=0.885\times 175\\V_{NaOH}=1032.5mL$