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

To neutralize different samples of HCl with 0.150 M NaOH, we calculate the number of moles of HCl in each sample and then determine the corresponding volume of NaOH solution needed. The volumes for the given samples of HCl are 25 mL, 20.17 mL, and 1032.5 mL.

Step-by-step explanation:

Determining the Volume of NaOH Solution Required for Neutralization

If we want to neutralize a sample of hydrochloric acid with sodium hydroxide (NaOH), we use the concept of molarity and the balanced chemical equation NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq). According to the balanced equation, the neutralization reaction occurs in a 1:1 molar ratio between NaOH and HCl. To find the volume of NaOH needed, we calculate the moles of HCl and then use the molarity of NaOH to find the corresponding volume.

• For 25 mL of a 0.150 M HCl solution: Number of moles of HCl = 0.025 L × 0.150 M = 0.00375 mol. The volume of 0.150 M NaOH required = 0.00375 mol × (1 L/0.150 M) = 0.025 L or 25 mL.
• For 55 mL of a 0.055 M HCl solution: Number of moles of HCl = 0.055 L × 0.055 M = 0.003025 mol. The volume of 0.150 M NaOH required = 0.003025 mol × (1 L/0.150 M) = 0.02017 L or 20.17 mL.
• For 175 mL of a 0.885 M HCl solution: Number of moles of HCl = 0.175 L × 0.885 M = 0.154875 mol. The volume of 0.150 M NaOH required = 0.154875 mol × (1 L/0.150 M) = 1.0325 L or 1032.5 mL.

Thus, the volumes of 0.150 M NaOH solution required to neutralize the given samples of HCl are 25 mL, 20.17 mL, and 1032.5 mL, respectively.

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)* V_(NaOH)=M_(HCl)* 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* V_(NaOH)=0.150* 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* V_(NaOH)=0.055* 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* V_(NaOH)=0.885* 175\\V_(NaOH)=1032.5mL`