# What is the pH of 0.256 M ammonium chloride NH4Cl?

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What is the pH of 0.256 M ammonium chloride NH4Cl? Kb NH3 ...

Source(s): ph 0 256 ammonium chloride nh4cl kb nh3 1 8x10 5: https://shortly.im/IyssV . ... Calculate pH of a buffer solution with 1.0 M ammonia ...

Other solutions

Ammonium ions act as an acid and transfer protons to water: NH4+ + H2O <--> NH3 + H3O+ Ka for...

For this, we can use the Henderson-Hasselbalch Equation, which states that: pH = pKa + log([base]/[acid...

The Henderson-Hasselbalch equation for buffers is pH = pKa + log (moles base / moles acid) In this problem...

Chem equilibrium qns?

How many grams of ammonium chloride (NH4Cl) would have to be dissolved in 500 mL of 0.20 M NH3 to prepare a solution having a pH of 10.00? Given: pKb of NH3 is 4.74. i could find out pKa, Ka and Kb from the given pKb, but then I don't know how to continue...

Your question seems to me merely interlocutory... and in no correct category. If I answered what you...

Nitrogen does require 3 electrons for an octet. That will show that it has 3 bond pairs and one lone...

What is the value of ∆H˚(J) for ammonium chloride? NH3(g)+HCl(g)––>NH4Cl(s)?

NH3(g)+HCl(g)––>NH4Cl(s) given the thermodynamically data of 298˚K ∆Hf˚(kJ/mol) S˚(J/K•mol) NH3(g): -46.19 192.5 HCl(g): -92.30 186.69 NH4Cl(s): -314.4 94.6

dH = (sum of reactants) - (sum of products) All ratios are 1/1 and moles cancel so your answer will...

What is the value of ∆G˚(J/mol) for ammonium chloride? NH3(g)+HCl(g)––>NH4Cl(s)?

given the thermodynamically data of 298˚K ∆Hf˚(kJ/mol) S˚(J/K•mol) NH3(g): -46.19 192.5 HCl(g): -92.30 186.69 NH4Cl(s): -314.4 94.6

do products - reactants to get delta H and delta S plug into delta G= delta H -T delta S make sure you...

What products are formed when you add NH4Cl (Ammonium Chloride) to these reactions?

I added Na3PO4 to MgSO4 in a test tube which resulted in this equation: 3MgSO4 + 2Na3PO4 ------> Mg3(PO4)2 + 3Na2SO4 Then I added NH4Cl to the test tube and I got a cloudy ...show more

I believe you should have gotten a cloudy precipitate in both cases before you added NH4Cl, since both...

NaOH + NH4Cl --> NaCl + H2O + NH3 0.025L x 2M = 0.05moles NaOH 0.05moles NaOH requires 0.05moles...

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