Calculate the pH of a solution that contains 7.8 x 10^-6 M OH^- at 25 degree C.

Question

Calculate the pH of a solution that contains 7.8 x 10^-6 M OH^- at 25 degree C. What is the pH of a solution whose pOH is 4.3 Determine the pH of a 0.265 M HCIO solution. The K_a of HCIO is 2.9 x 10 ^-8. 0.273 M solution of unknown Acid X in water has a pH of 1.25. What is the K_a of Acid X A student completely dissolves 31.5 mg of nitric acid (HNO_3) in water to result in a total of 22 m_ of solution. What is the expected pH of the resulting solution If 9.24 g of sodium fluoride (NaF) are completely dissolved in water to result in a total of 1.0 L of solution, what is the expected pH of this solution In a triprotic acid, which Ka has the highest value K_a1 K_a2 K_b3 Which one of the following salts, when dissolved in water, produces the solution with a pH closest to 7.00 NH4Br CaO KHSO_4 Csl

Explanation / Answer

5.5 pOH= -log (7.8*10-6)=5.1, pH= 14-pOH= 14-5.1=8.9

5.6 given pOH= 4.3, pH= 14-pOH= 14-4.3= 9.7

5.7

HClO-à H+ +ClO-

Ka= [H+] [CLO-]/ [HClO]

Let x= drop in concentration of HClO to reach equilibrium

[H+] =[CLO-] =x

2.9*10-6= x2/(0.265-x)

This quadratic equation can be solved using solver of excel and x= 0.000881M

pH= -log(0.000881)=3.055

5.8 pH =1.25

[H+] =10(-1.25)=0.056234

Let the acid be represented as HA

HA-à H+ +A-

[H+] =[A-]=0.056234

At Equilibrium

[HA] =0.273-0.056234=0.217

Ka= [H+] [A-]/ [HA] = (0.056234)2/ 0.217 =0.0146

5.9

Since HNO3 is strong acid,

Moalrity of HNO3 solution = moles of HNO3/ Volume

Volume of solution =22ml =22/1000=0.022L

Moles of HNO3= mass/Molecular weight= 31.5*10-3 g/63=0.0005

Molarity= 0.0005/0.022=0.0227 M

Since HNO3 completely ionizes HNO3à H+ +NO3-

[H+] =0.0227, pH= -log(0.0227)=1.64

5.10 Moles of NaF= 9.24/42=0.22moles. This is dissolved in 1L

NaF is formed from HF and NaOH (weak acid and strong base)

HF+NaOH-à NaF+ H2O

Molarity= 0.22M

Ka for HF= 7.2*10-4

Ka= [H+] [F-]/[HF]

Let x= drop in concentration of HF to reach equilibrium

Hence at equilibrium [H+]=[F-]=x and [HF] =0.22-x

Therefore= x2/(0.22-x)= 7.2*10-4

This equation when solved using solver gives x= .0122

[H+] =0.0122 pH= -log(0.0122)=1.91

5.11 Ka1 will be higher

5.11 Neutral salt when dissolved in water will have pH=7

1.KHSO4 is formed form KOH (Strong base) and H2SO4 (strong acid) and hence pH=7

2. NH4Br is formed from weak base (NH3) and strong acid (HBr). Hence pH not equal to 7

3.CaO is acidic and hence pH not equal to 7

4.CSI is neutrl because it is formed from strong base CSOH and strong acid (HI). hence neutral

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