A certain weak base has a Kb of 8.50 × 10-7. What concentration of this base will produce a pH of 10.05?
pH = 10.05 pOH = 3.95 [OH-] = 10^-pOH = 10^-3.95 = 0.00011220 M B + H2O <==> HB+ + OH- Kb = ([HB+] [OH-]) / [B] 8.50 x 10^-7 = [(1.122 x 10^-4) (1.122 x 10^-4)] / x x = 0.0148 M
A certain weak base has a Kb of 8.70 × 10-7. What concentration of this base will produce a pH of 10.25?
A certain weak base has a Kb of 8.70 × 10-7. What concentration of this base will produce a pH of 10.25? I tried to solve this but I was incorrect. Can you explain to me how to solve this problem? I want to know what I'm doing wrong. Best explanation gets 10 points. Thank you! My calculations: B + H2O----> BH + OH- Since pH=10.25, Total p=14 so 14-10.25=3.75 is pOH Concentration of pH= 10^ -10.25= 5.62X10^ -11 Concentration of pOH= 10^ -3.75= 1.78X 10^ -4 Kb=(BH)(OH)/(B)-----> 8.70X10^ -7= (5.62X10^ -11)(1.78X 10^ -4)/X----> X= (5.62X10^ -11)(1.78X 10^ -4)/8.70X10^ -7-----> X=1.15X10^ -8
A certain weak base has a Kb of 7.50 × 10-7. What concentration of this base will produce a pH of 10.39?
Method 1 : Denote the concentration of the base as B. pH at eqm = 10.39 pOH at eqm = pKw - pH = 14.00 - 10.39 = 3.61 [OH⁻] at eqm = 10⁻³˙⁶¹ M _____________ B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq) …. Kb = 7.50 × 10⁻⁷ Initial ________ y M __________ 0 M ____ 0 M Change ______ +10⁻³˙⁶¹ M __ -10⁻³˙⁶¹ M _ -10⁻³˙⁶¹ M At eqm ____ (y+10⁻³˙⁶¹) M __ 10⁻³˙⁶¹ M _ 10⁻³˙⁶¹ M Since Kb is very small, we can assume that 10⁻³˙⁶¹ ≪ y, i.e. [B] at eqm = (y+10⁻³˙⁶¹) M ≈ y M Kb = [BH⁺] [OH⁻] / [B] (10⁻³˙⁶¹)² / [B] = 7.50 × 10⁻⁷ [B] = (10⁻³˙⁶¹)² / (7.50 × 10⁻⁷) M = 0.0803 M ==== Method 2 : Denote the concentration of the base as B. pH at eqm = 10.39 pOH at eqm = pKw - pH = 14.00 - 10.39 = 3.61 [OH⁻] at eqm = 10⁻³˙⁶¹ M pOH = 0.5 * {pKb - log[B]} 0.5 * {-log(7.50 × 10⁻⁷) - log[B]} = 3.61 log(7.50 × 10⁻⁷) + log[B] = -7.22 log[B] = -7.22 - log(7.50 × 10⁻⁷) [B] = 10^{-7.22 - log(7.50 × 10⁻⁷)} M = 0.0803 M
A certain weak base has a Kb of 8.3 X 10^-7. What concentration of this base will produce a pH of 10.33?
A- + H2O <=> HA + OH- Kb = [HA].[OH-] / [A-] Assuming [HA] = [OH-] Kb = [OH-]² / [A-] [OH-]² = Kw² / [H3O+]² where [H3O+] = 10^-10.33 and Kw = 1 x10^-14 8.3 x10^-7[H3O+]² / Kw² = 1 / [A-] [A-] = 5.51 x10^-2 mol/L
A certain weak base has a Kb of 8.90 × 10-7. What concentration of this base will produce a pH of 10.17? help?
pH = 10.17 pOH = 3.83 [OH-] = 10^-pOH = 10^-3.83 = 1.4791 x 10^-4 M B + H2O <==> HB+ + OH- Kb = ([HB+] [OH-]) / [B] 8.90 x 10^-7 = (1.4791 x 10^-4) (1.4791 x 10^-4) / x x = 0.0246 M
A certain weak base has a Kb of 8.20 x 10^-7. What concentration of this base will produce a pH of 10.25?
10^(-10.25) = 5.6*10^-11= [H+] (5.6*10^-11)(oh-) = 1.0*10^-14 OH- = 1.79*10^-4
A certain weak base has a Kb of 7.40 × 10-7. What concentration of this base will produce a pH of 10.39?
Kb = 7.4x10^-7 pH = 10.39 ? [base] B^- + H2O ---> HB + OH^- Kb = [HB][OH]/[B-] For pH of 10.39 the pOH will be 3.61 and thus the [OH] = 2.45x10^-4 Then, 7.4x10^-7 = (2.45x10^-4)^2/[B-] [B-] = 6x10^-8 / 7.4x10^-7 [B] = 0.081 M
A certain weak base has a Kb of 8.60 × 10-7. What concentration of this base will produce a pH of 10.34?
The equilibrium reaction is below B + H2O <=> CA + HO- Equilibrium equation: Kb= [HO-][CA]/[B] [B]= concentration of base [CA]= concentration of conjugate acid Also [HO-]= [CA] = x 14 - pH = pOH............................14-10.34 = 3.66 -log[HO-] = pOH................................... -log[HO-]= 3.66 [HO-] = 10^ (-pOH) .............................. [HO-] = 10^-3.66 = 4.786 x 10^-8 Since [HO-]= [CA] = x and we now know that x is 4.786 x 10^-8 So using the equilibrium equation and replacing x for [HO-] and [CA] we get: Kb = x^2/[B] Solve for [B] [B] = x^2/Kb [B] = (4.786 x 10^-8)^2 / (8.60 × 10-7) [B]= 0.0557 Hope it helps!
A certain weak base has a Kb of 9.00 × 10-7. What concentration of this base will produce a pH of 10.28?
pOH = 14 - pH pOH = - log [OH^-] 14 - pH = - ½ log (Kb M) 14 - 10.28 = - ½ log (9.00 * 10^-7 * M) M = 0.040342 Molar