1. Calculate the pH of 0.072M HNO2 2. Calculate the pH of a solution made with 0

Question

1. Calculate the pH of 0.072M HNO2 2. Calculate the pH of a solution made with 0.072M HNO2 and 0.065M NaNO2 3. What effect to the pH did adding a salt have on the HNO2 solution? 4. Calculate the pH of the solution in #2 after the addition of 0.017 moles of NaOH 1. Calculate the pH of 0.072M HNO2 2. Calculate the pH of a solution made with 0.072M HNO2 and 0.065M NaNO2 3. What effect to the pH did adding a salt have on the HNO2 solution? 4. Calculate the pH of the solution in #2 after the addition of 0.017 moles of NaOH 2. Calculate the pH of a solution made with 0.072M HNO2 and 0.065M NaNO2 3. What effect to the pH did adding a salt have on the HNO2 solution? 4. Calculate the pH of the solution in #2 after the addition of 0.017 moles of NaOH

Explanation / Answer

1)

HNO2 = 0.072M

Ka of HNO2 = 4.5x10^-4

for weak acids

[H+] = sqaure root of KaxC

]H+] = sqaure root of (4.5x10^-4x0.072)

[H+= 0.569x10^-2M

-log[H+]= -log(0.569x10^-2)

PH= 2.24

b) HNO2 = 0.072M

NaNO2= 0.065M

Ka= 4.5x10^-4

-log(ka)= -log(4.5x10^-4)

PKa = 3.35

PH = PKa + log[salt]/[acid]

PH= 3.35+log(0.065/0.072)

PH= 3.31

c) by adding the salt , PH opf the solution is decreases.

d) number of moles of HNO2 = 0.072Mx1L= 0.072 moles

number of moles of NaNO2 = 0.065Mx1L= 0.065 moles

number of mole pf NaOH= 0.017 moles

after addition of NaOH

number of moles of HNO2 = 0.072 - 0.017 =0.055 moles

number of moles of NaNO2 = 0.065 +0.017=0.082 moles

PH= 3.35 + log(0.082/0.055)

PH=3.52

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