What mass of solid NaCH3CO2 should be added to 1.0 L of 0.50 M CH3CO2H to make a buffer with a pH of 7.21?
Use Henderson -Hasselbach eqn HAc=CH3COOH and [Ac-]= CH3COO- pH =pKa + log [Ac-]/[HAc] 7.21=7.21+log [Ac-]/0.50 log [Ac-]/0.5=0 [Ac-]/0.5=1 [Ac-]= 0.5 Molarity of NaCH3CO2= 0.5 moles =0.5 Mass= molesX molar mass = 0.5x 82=41.0g
What is the molar solubility of Zn(OH)2 in a solution buffered at pH of 8.0? (Zn(OH)2, Ksp = 3.0 × 10 −17)?
3 x 10-5 Zn(OH)2 <..........> Zn + 2 OH 3 x 10-17 = [Zn] x [OH]2 The OH concentration in a solution with a pH of 8 is 1 x 10-6. The concentration of OH ions contributing to the total OH ion concentration that is added by the ionization of zinc hydroxide is negligible, especially since its Ksp is so small, so they can be ignored. This reduces your equation to: 3 x 10-17 = [Zn] x [1 x 10-6]2 which comes down to: 3 x 10-17 / 1 x 10-12 = [Zn] which equals 3 x 10-5 Since the concentration of the zinc ions is the same as the concentration of zinc hydroxide, (all the zinc ions came from the dissolved zinc hydroxide), the solution contains 3 x 10-5 moles per liter of zinc hydroxide.
Calculate the molar solubility of calcium fluoride, CaF2, in a 0.050 M NaF solution. ?
let x = mol/L of CaF2 that dissolve we get x mol/L Ca2+ and 2x mol/L F- we have arleady 0.050 M of F- so [F-] = 2x + 0.050 Ksp = [Ca2+][F-]^2 3.9 x 10^-11 = (x)( 2x+ 0.050)^2 2x is small compared to 0.050 3.9 x 10^-11 = (x) ( 0.050)^2 x = molar solubility = 1.6 x 10^-8 M
How many moles and how many grams of sodium chloride are present in 250 cm3 of 0.5 molar NaCl solution?
The number of moles of a solute (NaCl in this case) is equal to the concentration multiplied by the volume (in L).ie. n = c x vSo the number of moles of NaCl is 0.5 x 0.25 = 0.125 molesNow, the mass of a solute equals the number of moles multiplied by its molar mass.ie. mass = n x MMSo the mass of NaCl = 0.125 x (22.9898 + 35.453) = 7.31 gI hope this helps
Chemistry help !!! solubility equilibrium?
Yes, there is a method. What I will do is calculate the calcium and fluoride concentrations created by the dissolving, then put them into the Ksp expression. We will then compare that answer, called Qsp, to the Ksp. MV = mass / molar mass (x) (100 L) = 7.8 g / 78.074 g/mol M = 0.000999 M ---> 0.0010 M [Ca^2+] = 0.0010 M [F^-] = 0.0020 M Qsp = (0.0010) (0.0020)^2 = 4.0 x 10^-9 Qsp is larger than Ksp ---> we conclude that not all the CaF2 will dissolve because you cannot exceed the Ksp (given constant temperature). Comment: I phrased it the way I did because formation of a precipitate was not involved. Formation of a precipitate happens, in the most common situation, where two solutions of soluble reagents are mixed and some solid forms. This is a precipitate forming. In the above scenario, we throw the CaF2 solid into the 100 L and stir. Not all of the 7.8 g dissolves and some solid sinks to the bottom. This is not the formation of a precipitate.
How many milliliters of 0.025 M H2SO4 are required to neutralize exactly 525 ml of 0.06 M KOH?
Let us first find out how much H2SO4 is required to neutralize 525 ml of 0.06 M KOH.Why ? Because 1 molecule of H2SO4 gives 2 H+ ions per molecule while only one H+ ion is required to neutralize 1 molecule of KOH. So, 1 molecule of H2SO4 can neutralize 2 molecules of KOH. Hence, we would require 525 ml of 0.03 M H2SO4 to neutralize 525 ml of 0.06 M KOH. How will we prepare 525 ml of 0.03 M H2SO4 ?Now, we have 0.025 M H2SO4 and we do not know how much volume we have.We will use the standard N1 X V1 = N2 X V2 for this calculation.N1=0.025 M; V1=unknown; N2=0.03 M and V2=525 ml.So V1= (0.03 X 525)/(0.025) = 630 ml.
How do I calculate the solubility of Fe(OH)3?
I noticed there was no Ksp value included and temperature. But we can try to calculate it.I would start off by calculating the pH of the buffer. Use the Henderson Hasselbeck equation.pH=pka+log [A-]/[HA]ka (NH4+)=5.6*10^-10=-log(5.6*10^-10)=9.25pH=9.25 +log 2/2=9.25(quite alkaline in nature)Write out you equation and calculate kspIf they gave you a specific ksp value use it, otherwise I calculated it.Fe(OH)3 —- Fe3+ =3OH-Ratio=1:3[OH-]=3Fe3+=3*4.7*10^-6MKsp=[Fe3+][OH-]3=1.3175*10^-20MNow, you can calculate the molar solubility. remember to convert ml to LFe(OH)3 <> Fe3+ + 3OH-Ksp= ( Fe3+)(OH-)^3When x moles/L Fe(OH)3 dissolve we get x moles/L Fe3+ and 3x moles/L OH-Substitute these in Ksp expression1.3175*10^-20 = x ( 3x)^3solve for x to get your solutionEdit: I forgot to include the pH in here:pOH=4.5[OH-]=10^-4.5=3.16227*10^-5Now it will look like this:Ksp = x ( x + 3.16227*10^-5)^3 solve for xx=molar solubility
What is the meaning of 20% NaOH solution?
What is the meaning of 20% by mass of NaOH solution?20% by mass means 20 g of the NaOH is dissolved in 100 g of solution.What is the meaning of 20% mass by volume solution?20% mass by volume means 20 g of the NaOH is dissolved in 100 mL of the solution.Depends on the concentration terms meaning and values also changes.You may interest to know more about it, please visit: www.organichemistryguide.com
How do you prepare a 1M NaOH solution?
NaOH is a secondary slandered,it can absorb moisture from air .so we can't prepare it's standard solution by direct weighing of NaOH.A)first preparation of approximately 1 M NaOH solution:—Molarity=no of moles÷volume in Lif you want to prepare approx 1M NaOH solution in 250ml volumetric flask then1000ml—1M—40g of NaOH250ml. —1M — x g of NaOHForm above equation1000ml×1M×X g=250ml×1M×40gX=[250×1×40]÷[1000×1]X=10g of NaOHThat is to prepare 250 ml NaOH solution dissolve 10 g NaOH crystals in 250 ml volumetric flask and dilute up to the mark with the distilled water you get approx 1M NaOH solution.B) Standardization of NaOH solution:—Take approx 1M NaOH ,25 ml in a conical flask add 2–3 drops of phenolphthalein indicator and titrate with standard oxalic acid solutionApprox1M NaOH Vs oxalic acidM1V1. = M2V2M1×25= 1×V ml acid require for neutriM1. = Vml÷25 = X MThis M1 is exact molarity of prepared NaOH solution