calculate the volume of .1M HCl that is required to react with 50 mL of .1M ammonia to form a mixture that has pH 9
Use the Henderson-Hasselbalch equation but this one takes a little more work than the last one.
Write the equation.
NH3 + HCl ==> NH4Cl
You start with 50 mL x 0.1 M = 5 mmoles and you want to add HCl to it. The easy way to do it is to determine what the ratio Base/acid must be and go from there.
2 C2H2 + 5O2 >>> 4CO2 + 2H2O
In the above equation . How many grams of CO2 are produced when 109.2 grams of C2H2
To calculate the volume of 0.1 M HCl required to react with 50 mL of 0.1 M ammonia to form a mixture with pH 9, you need to consider the stoichiometry of the reaction and the acid-base equilibrium involved.
The reaction between HCl and ammonia (NH3) can be represented as:
HCl + NH3 → NH4Cl
This reaction produces ammonium chloride (NH4Cl) when HCl reacts with ammonia.
Now, let's consider the acid-base equilibrium in the mixture. Ammonia acts as a base and reacts with water to produce ammonium ions (NH4+) and hydroxide ions (OH-):
NH3 + H2O ⇌ NH4+ + OH-
The pH of a solution is a measure of the concentration of hydrogen ions (H+). A solution with pH 9 indicates a low concentration of H+ and a high concentration of OH- ions. Since HCl is a strong acid that completely dissociates in water, it will provide H+ ions in the reaction.
To neutralize the OH- ions in the mixture and achieve a pH of 9, an equal amount of H+ ions is required. Therefore, the amount of H+ ions provided by HCl must be equal to the amount of OH- ions present in the solution.
Given that the HCl concentration is 0.1 M and the volume of ammonia is 50 mL (0.05 L), we can set up the following equation based on the stoichiometry and acid-base equilibrium:
0.1 M HCl × V HCl = 0.1 M NH3 × 0.05 L
To solve for V HCl (the volume of HCl required), we rearrange the equation:
V HCl = (0.1 M NH3 × 0.05 L) / 0.1 M HCl
V HCl = 0.05 L
Therefore, the volume of 0.1 M HCl required to react with 50 mL of 0.1 M ammonia to form a mixture with pH 9 is 0.05 L (50 mL).