Calculating pH of CH3NH2 and CH3NH3+Cl- Solution

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To calculate the pH of a solution containing 12.4g of CH3NH2 and 13.5g of CH3NH3+Cl-, the Henderson-Hasselbalch equation is applied, where pH = pKa + log [salt]/[acid]. The molar masses are used to find the moles of each component, resulting in 0.199 moles of the salt (methylammonium chloride) and 0.399 moles of the acid (methylamine). The molarity is calculated as 0.099M for the salt and 0.199M for the acid in a 2.00 dm3 solution. Substituting these values into the equation gives a final pH of 10.5. The discussion emphasizes the importance of understanding concentration in relation to the Henderson-Hasselbalch equation.
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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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