Problem: 1.00 g CH3NH3Cl (MW = 67.52 g/mol) and 1.00 g Ba(OH) 2∙8H2O (a strong base, MW = 315.47 g/mol) are added to 80 mL of water and dissolved. The resulting solution is diluted to 100 mL with water at 25°C.Calculate the pH of the solution (pKa(CH3NH3+) = 10.62).A. 11.00B. 10.49C. 10.00D. 11.17E. 10.75

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We are being asked to calculate the pH of a solution prepared by dissolving 1.00 g CH3NH3Cl (MW = 67.52 g/mol) and 1.00 g Ba(OH) 2∙8H2O (a strong base, MW = 315.47 g/mol) to 80 mL of water. The resulting solution is diluted to 100 mL with water at 25°C.


We will calculate the pH of the solution at the equivalence point using the following steps:

Step 1. Write the chemical equation for the reaction between CH3NH3Cl and Ba(OH)2.
Step 2. Calculate the initial amounts of CH3NH3Cl and Ba(OH)2 in moles before the reaction happens.
Step 3. Construct an ICF Chart.
Step 4. Calculate pH.


Step 1. Write the chemical equation for the reaction between CH3NH3Cl and Ba(OH)2.

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Problem Details

1.00 g CH3NH3Cl (MW = 67.52 g/mol) and 1.00 g Ba(OH) 2∙8H2O (a strong base, MW = 315.47 g/mol) are added to 80 mL of water and dissolved. The resulting solution is diluted to 100 mL with water at 25°C.

Calculate the pH of the solution (pKa(CH3NH3+) = 10.62).

A. 11.00

B. 10.49

C. 10.00

D. 11.17

E. 10.75

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What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Weak Acid Strong Base Titrations concept. You can view video lessons to learn Weak Acid Strong Base Titrations. Or if you need more Weak Acid Strong Base Titrations practice, you can also practice Weak Acid Strong Base Titrations practice problems.