Calculating pH of CH3NH2 and CH3NH3+Cl- Solution

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Discussion Overview

The discussion revolves around calculating the pH of a solution containing CH3NH2 (methylamine) and CH3NH3+Cl- (methylammonium chloride). Participants explore the application of the Henderson-Hasselbalch equation in this context, focusing on the necessary calculations for molarity and pH determination.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • Joe presents a problem involving the calculation of pH for a specific solution and provides the weights of the compounds involved.
  • One participant suggests using the Henderson-Hasselbalch equation, providing a step-by-step calculation of moles and molarity for both the acid and salt.
  • Another participant comments that the volume of the solution can be omitted in the calculations since both components are in the same volume, implying that the mole ratio is sufficient for the pH calculation.
  • A later reply indicates that omitting the volume does not affect the final pH result, emphasizing the purpose of the explanation regarding the Henderson-Hasselbalch equation.

Areas of Agreement / Disagreement

Participants generally agree on the use of the Henderson-Hasselbalch equation for this calculation, but there is a minor disagreement regarding the necessity of including volume in the calculations.

Contextual Notes

Participants do not explicitly resolve the implications of omitting volume in the calculations, leaving some uncertainty regarding its impact on the final pH value.

josephcollins
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Hi ppl, I got a problem with this question

Calculate the pH of 2.00dm3 of an aqueous solution which contains 12.4g of CH3NH2 and 13.5g of methylammonium chloride, CH3NH3+Cl-. The Pka of methylammonium chloride is given as 10.8


Thanks a lot for any help,
Joe
 
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pH of a solution

Hi
we will apply Henderson-Hasselbach equation here
which says that pH = pKa + log [salt] / [acid]
salt is CH3NH3Cl and acid is CH3NH2
to determine concentartion we have to calculate the Molarity of each.
mol wt of salt = 67.521, mol wt of acid = 31.068
therefore moles of salt = wt / mol wt = 13.5/67.521 = 0.199
moles of acid = 12.4 / 31.068 = 0.399
as the solution is 2dm3 that means 2 litres, hence molarity of each will be
salt = 0.199 / 2 = 0.099M, acid = 0.399 / 2 = 0.199M
putting the values in equation
pH = 10.8 + log 0.099 / 0.199
pH = 10.5
 
Your approach seems to be okay, but you can omit the volume as they are present in the same 2 liters of solution, so just divide the mole values of base and acid each.
 
pH

any how it won't effect the answer... it was just to explain the basis of henderson hasselbach equation and what exactly means by concentration so it can be used in varying occasions
 

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