This article talks about does any solid cu(oh)2 form when 0.075 g of Koh is dissolved in 1.0 l of 8m naNO3? information related to it, including what the chemical equation would be if you were to dissolve 0.075g of KOH in 1 liter of NaNO3 with an ionic concentration of 8M and molecular weight for CuO NO3 that is 223.871 grams per mole, as well as why this reaction might not have a solution that contains only CuO (solid).
The chemical equation would be Cu(OH) ̄NO ̇ (solid), Na(sodium hydroxide solution). The reaction does not have a solid solution because it is an acid-base reaction. In order to solve the problem, we could dissolve our KOH in water instead of NaNO and then add that to our NO. This will result in a neutral pH so that there is no product formed with OH-. It also means you need more solvent than before, which increases your volume requirement by four times.
This is an article about does any solid cu(oh)20 form when 0.075 g of Koh is dissolved in 1000 ml of naNO30? information related to it such as mole and the reason for its insoluble product being formed with OH-. It also includes reasons we would need more solvent than before like water even if NaOH and NO are mixed together because they will be neutralized by hydrogen ions at their pH balance which makes them unproductive.
This information relates to ;
do any solids CUO dissolve with a solution containing KOH (potassium hydroxide)? The reason for it being insoluble in water is that OH- ion needs to be present so these two ions don’t fixate each other as products while reacting with NO.
As mentioned above, the difference between hydrochloric acid and nitric acid is their concentrations of hydrogen (H+) ion concentration. Hydrochloric Acids has an H+ concentration of 0.06M and Nitric Acid’s H+ concentration is around 0.02M at 25°C or 50% relative humidity.”
The major difference between these acids is actually how they react when combined together. This is because hydrochloric acid and nitric acid have different charges.
This means the reaction will not happen, therefore no solid CuO forms.”
Therefore, do any solids CUO form when 0.075g of KOH are dissolved into l at 25°C? No! There won’t be any solids CUO formed after adding KOH to NaNOs solution because this process can only produce soluble compounds such as Cu(OH)SO. The major difference between these acids is actually how they react when combined together. This is because hydrochloric acid and nitric acid have a different charge.”
Hydrochloric acid is a strong acid which has a charge of -0.02 and nitric acid is a weak acidic with a charge around 0.01.”
“Hydrochloric Acid dissociates into the ions H+ and Cl- when it reacts with water, while Nitric Acid does not separate in this way because there are no charges to be neutralized by combining them together.”
This means that hydrochloric acids will react much more aggressively than nitric acids, just as its stronger concentration would lead you to believe. The higher the concentrations of hydrogen ion (H+) is relative to hydroxide ion (OH-) becomes more soluble these compounds become until they reach an equilibrium where the solubility levels off at pK values.
In order to answer the question of does any solid Cu(OH)₂ forms when 0.075gof Koh is dissolved in one liter of NaNO₃, we would need to use an equation that helps determine solubility and equilibrium. The most common equations for this are Ksp and Kw which stand for soluble species product (K sp ) and water-soluble species equilibrium constant respectively (kW). These two constants can help us find out how much copper hydroxide will dissolve with time at different concentrations:
Cu²+(aq)+HCOONa+ ⇄ CO32-(aq) + OH-(aq)
The K w value is -log [CO32-]/[HCOONa], so the equilibrium constant expression is -log (Kw) = log K w + log
(Cu²+). For instance, if we have a solution with Cu²+(aq) = 0.075g, HCOONa=0.107mol∙L− and [OH-(aq)]=0.009 mol ∙ L − then:
log (Kw)= (-login K w ) + (-log in Cu²+)
log Kw=-(-11)-(-11)+22=12
Therefore, the answer to does any solid cu(oh)₃ form when .07 g of Koh is dissolved in 0.01 l of .0001 × 104 mol∙L− is no, as there will be an increase in K w
