How many moles of NH3 (Kb= 1.8 × 10‒5) must be dissolved in water in order...

How many moles of NH3 (Kb= 1.8 × 10‒5) must be dissolved in water in order to prepare 0.616 L of solution with a pH of 11.22?

Expert Solution

queen_honey_blossom answered 2 years ago

Take 12.5 mL of 0.200 moles of aqueous Ammonia (NH3), whose Kb = 1.8 x 10-5,...

Take 12.5 mL of 0.200 moles of aqueous Ammonia (NH3), whose Kb = 1.8 x 10-5, and Titrate with 0.100 mol HCl, calculate PH at 5 mL HCl, 10 mL , 15 mL, 20 mL, 24.5mL, 25mL, 25.5mL, 30mL 40.0 mL, abd 50 mL HCl. Graph the solution with hoizontal axis being the PH up to 14 and The y-axis being mL HCl.

Given that Ka for HCN is 4.9×10−10 and Kb for NH3 is 1.8×10−5 , calculate Kb...

Given that Ka for HCN is 4.9×10−10 and Kb for NH3 is 1.8×10−5 , calculate Kb for CN− and Ka for NH4+ . Enter the Kb value for CN− followed by the Ka value for NH4+ , separated by a comma, using two significant figures.

In a titration of a 250mL solution of 0.05M ammonia (NH3) (Kb=1.8 * 10^-5) with a...

In a titration of a 250mL solution of 0.05M ammonia (NH3) (Kb=1.8 * 10^-5) with a concentrated (0.2M) solution of HCL, calculate the following: 1. What is the initial pH before adding HCL? 2. What is the pH of the system after adding 20mL of the 0.2 M HCL? 3. What is the pH at equivalence? *I have the answers, I really need to know how to do the work to get to the answers.*

A buffer is made by combining 5.55 g of NH3 (Kb = 1.8 X 10-5) with...

A buffer is made by combining 5.55 g of NH3 (Kb = 1.8 X 10-5) with 4.78 g of HCl and diluting to a volume of 750.0 mL. a. What is the pH? b. What is the pH after adding 60.0 mL of 2.00 M HBr(aq)? c. What is the pH after adding 120.0 mL of 2.00 M HBr(aq)?

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How many moles of NH3 (Kb= 1.8 × 10‒5) must be dissolved in water in order...

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Expert Solution

Related Solutions

Take 12.5 mL of 0.200 moles of aqueous Ammonia (NH3), whose Kb = 1.8 x 10-5,...

Given that Ka for HCN is 4.9×10−10 and Kb for NH3 is 1.8×10−5 , calculate Kb...

In a titration of a 250mL solution of 0.05M ammonia (NH3) (Kb=1.8 * 10^-5) with a...

A buffer is made by combining 5.55 g of NH3 (Kb = 1.8 X 10-5) with...

NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts...

NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts...

NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts...

NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts...

NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts...

NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts...