If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
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What is base dissociation constant?
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The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 2.8× 10^(-9)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{2.8×10^(-9) }
= 3.5× 10^(-6)
Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
DISCLAIMER: The above question have mistake. The correct question is given as
Question:
Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?
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Answer:
38.9 grams of 
Explanation:
0.187 mol BaCl2 x 
0.187 m x 208 g/m
0.187 x 208 g
38.896 g --> 38.9 g BaCl2
Answer:
1.395J/g°C
Explanation:
The following were obtained from the question:
Q = 6527J
M = 312g
ΔT = 15°C
C =?
Q = MCΔT
C = Q/MΔT
C = 6527/(312 x 15)
C = 1.395J/g°C
The specific heat capacity of the substance is 1.395J/g°C
Answer:
45 neutrons
Explanation:
Bromine has 35 protons and a mass number of 80. a) How many neutrons does the atom of bromine have? The mass number = protons + neutrons. Bromine has a mass number of 80 and 35 protons so 80-35 = 45 neutrons.
Answer:
pH = 2.46
Explanation:
Hello there!
In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:
Whereas the moles of the salt are computed as shown below:
So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:
![[salt]=0.01428mol/0.0276L=0.517M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D0.01428mol%2F0.0276L%3D0.517M)
Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:
Whose equilibrium expression is:
![Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BC_6H_5NH_2%5D%5BH_3O%5E%2B%5D%7D%7BC_6H_5NH_3%5E%2B%7D)
Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:
Whereas x is:
Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:
Regards!
