What is the molarity of the HCl solution? Help Solving Dilution & Moles Problems

  • Context: Chemistry 
  • Thread starter Thread starter chmdummy1982
  • Start date Start date
  • Tags Tags
    Dilution Moles
Click For Summary

Discussion Overview

The discussion revolves around solving dilution and moles problems related to a hydrochloric acid (HCl) solution and sodium hydroxide (NaOH) reactions. Participants are attempting to clarify their understanding of the dilution formula and the stoichiometry involved in the reactions.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant attempts to use the dilution relationship (Mi x Vi = Mf x Vf) to calculate the volume of 0.500 M NaOH needed for a 500 mL solution of 0.250 M NaOH, expressing uncertainty about their calculations.
  • Another participant reminds that Molarity (M) is defined as moles per liter and suggests using algebra to find the unknown initial volume while keeping units consistent.
  • A different participant asks for clarification on the second problem, questioning what specific information is being sought and emphasizing the importance of balanced equations and reactant ratios.
  • One participant notes that the second question lacks the molarity of the hydrochloric acid, implying that this is a critical piece of information needed to proceed.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the solutions to the problems, and multiple viewpoints regarding the approach to solving the problems remain. There is uncertainty about the specifics of the second problem and the necessary information to solve it.

Contextual Notes

The first problem's calculations may contain errors, as indicated by the participant's own doubts. The second problem is incomplete without the molarity of HCl, which affects the ability to solve it.

Who May Find This Useful

This discussion may be useful for students or individuals seeking assistance with chemistry problems involving dilution, molarity, and stoichiometry in acid-base reactions.

chmdummy1982
Messages
2
Reaction score
0
I have two problems that I have trouble with. One I attempted, but I don't think I came up with the right formula. Any help would be appreciated!

Problem one

Use the dilution relationship (Mi x Vi = Mf x Vf) to calculate the volume of 0.500 M NaOH needed to prepare 500 mL of .250 M NaOH.

This is my work

? L Solution = 500mL x (1L/1000mL) = .500L

? mol NaOH = .500 L Solution x (0.250 M NaOH/1 L Solution) = .125 mol NaOH

? L solution = .125 mol NaOH x (1L Solution/.500 mol NaOH) = .25

The problem then asks me to round my answer to the nearest ones place, so i have a feeling this equation has an error. Can anyone guide me to getting it right? Thanks!



Problem 2
49.22 ml of a 2.01 M naOH solution reacts completely with 40.28 mL of HCl solution according to the blanaced chemical reaction shown below:

HCl (aq) + NaOH(aq) -> NaCl(aq) + H2O(I)
 
Physics news on Phys.org
Reminder: M means moles per liter, or moles per 1000 milliliters.
Your initial volume is "unknown" and your final volume is expected to be 500 ml.
In symbols as MILLILITERS, Vf=Vi+Vu, where Vu is unknown volume to add.

Simply substitute the values given and use simple algebra, but first keep units in moles per thousand milliliters for M, and milliliters for V.

Start with the concentration formula relationship for ease. Find the unknown initial volume. Use the simple volume addition relationship next.
 
chmdummy1982 said:
Problem 2
49.22 ml of a 2.01 M naOH solution reacts completely with 40.28 mL of HCl solution according to the blanaced chemical reaction shown below:

HCl (aq) + NaOH(aq) -> NaCl(aq) + H2O(I)
Heya,

So what is the question? To find the volume of products? Also, do you have any workings out? If not, what do you know about the balanced equations and ratios of reactants that will help?

The Bob
 
The second question is missing the molarity of the hydrochloric acid. This is what the question then must ask.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K