E, NH4+The base pairs v its conjugate acid. The basic is the proton acceptor, therefore the conjugate mountain the the substance that receives that proton.

You are watching: The conjugate acid of nh3 is ________.


B, OH-. The conjugate base is pared through the acid. The acid is the proton donor, so the conjugate basic is the one that becomes deprotinated.
The problem HClO4 is taken into consideration to it is in A, a weak acidb, a weak basec, a solid acidd, a solid base, E, a neutral compound
Acid stamin decreases in the series HI>HSO4->HF>HCN. I m sorry of these anions is the WEAKEST base?a, I-B, SO42-C, F-D, CN-
A, I-. Because HI is the the strongest acid, the dissociates completely in solution. Therefore, the anion is the weakest since it walk bye byes.
D, HBrOSo considering ours memorized list of strong acids, HBr is the strongest. Also, need to take into consideration that the more oxygens over there are, the more polar the becomes, make it more willing come dissociate, boosting the mountain strength. Therefore the weakest would need to be the one with the the very least Oxygens.
The mountain dissociation constant Ka=1.2610^-2 because that HSOr- and is 5.610^-10 for NH4+. I beg your pardon statement about the complying with equlibrium is correct?HSO4- (aq) + NH3(aq) SO42- (aq) + NH4+ (aq)a, the reactants will certainly be favored since ammonia is a more powerful base 보다 the sulfate anion.b, the products will it is in favored since the hydrogen sulfate ion is a more powerful acid than the ammonium ionc, no reactants or products will be favored because all of the species are weak mountain or basesd, the initial concentration of the hydrogen sulfate ion and also ammonia should be well-known before any kind of prediction deserve to be madee, this reactinon is impossible to predict, since the solid acid and also the weak base appear on the same side that the equation
b, the assets will be favored due to the fact that the hydrogen sulfate ion is the solid acid 보다 the ammonium ion. Ka=/ , so the greater the Ka, the stronger the acid. Therefore, since the Ka for hydrogen sulfate is much larger, the mountain is stronger and also will journey the reaction in the front direction, creating products.
which that the adhering to salts will type an acidic solution when liquified in water?a, NaClB, NaNO2C, NH4NO3D, NH4CN
C, NH4NO3since the no3 will type HNO3 i beg your pardon is a strong acid. No even similar to the weak base the is formed by NH3.
a systems is prepared by adding .1n that NaF come a 1.00L that water. I beg your pardon statement about the solution is correct?a, the solution is basicb, the systems is neutralc, the solution is acidicd, the concentrations of fluoride ions and sodium ions room equale, the concentrations of hydronium ions and also hydroxide ions space equal
what is the effect on the equilibrium once sodium formate (HCOONa) is included to a solution of formic acid (HCOOH)?HCOOH H+ + HCOO-a, over there is no readjust in eqb, the eq shifts to the rightc, an ext info is neededd, the equilibrium shifts to the lefte, the pH decreases
which that the adhering to results in a buffered solution?a, a soln containg a solid acid and also its conjugate baseb, a soln containing a strong base and also its conjugate acidc, a soln containing a catalystd, a equipment containing two organic liquidse, a solution containing a weak acid and also its conjugate base
e, a systems containing a weak acid and its conjugate base. (in similar amounts) this is the definition of a buffer.
when the concentration of the conjugate basic of a weak mountain is much less than the concentration that the weak acid in a buffer solution, the pH that the solution is a, less than the pKab, greater than the pKac, constantly acidic d, constantly basice, same to the pKa
a, much less than the pKapH=-logKa+log/say original pKa=7...-logKa+log/ create a negative, so when added to the -logka, that is much less than the original pH, therefore its less than pKa.
which pair of substances is capable of developing a buffer in an aqueous solution?a, H3PO4, Na3PO3b, HNO3, NaNO3C, HCl, NaCld, H2CO3, NaNO2E, CH3COOH, CH3COONa
E, CH3COOH, CH3COONaremember the definition of a buffer, weak acid and also its conjugate base. (or vise versa).
Calculate the pH the a buffer solution that consists of .25M benzoic acid and also .15M sodium benzoate. Ka=6.5E-5 because that benzoic acid.

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you have 500ml the a buffer equipment containing .20M CH3COOH and .3M CH3COONa. What will certainly the pH the this systems be after the addition of 20ml the 1MNaOH come solution? Ka(CH3COOH) = 1.8E-5
a 50 ml sample of .50 M HCl is titrated v .50M NaOH. What is the pH that the equipment after 28 ml that NaOH have been added to the acid?
a 20 ml sample of a .3MHBr is titrated through .15 M NaOH. What is the pH that the systems after the 40.3 mL of NaOH have been added to the acid?
when a weak mountain is titrated with a strong base, the pH at the equivalience allude is a, better than 7b, same to 7c, less than 7, d, equal to the pKa of the acide, is same to the pKb of the conjugate base
the molar solubility that magnesium lead carbonate (MgCO3) is 1.8E-4 n/l. What is the Ksp because that this compound?
which that the following compounds is appreciable more soluble in 1M HNO3 보다 in pure water?A, FeCO3B, AgBrC, BaSO4D, NaNO3E, NaCl
will a precipitate of magnesium flueoride MgF2 kind if 300ml the 1.1E-3 M MgCl2 is added to 500ml that 1.2E-3 M NaF? Ksp (MgF20=6.0E-9)a, yes since Q>Kspb, no since Qc, no due to the fact that Q=Kspd, yes due to the fact that Qe, no since Q> Ksp
consider the dissolution of MnS in water . MnS(s) + H2O (l) Mn2+ (aq) +HS- (aq) + OH- (aq)how is the solubility of manganese (II) sulfide influenced by the addition of aqueous potassium hydroxide (KOH) to the to the system?a, the solubility will certainly be unchangedb, the solubility will certainly decreasec, the solubility will increased, the quantity of KOH included must it is in known before its effect can be predictede, the pKa the the H2S is needed before a trustworthy prediction deserve to be made.
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