Classify each of the following solutions as acidic, basic or neutral: pH = 2.1 p

Question

Classify each of the following solutions as acidic, basic or neutral: pH = 2.1 pH = 7.0 pH = 9.2 pH = 6.8 Calculate the pH of a solution with [H_3O^+] = 1.0 times 10^-12 M. Calculate the pH of a solution with [H_3O^+] = 3.75 times 10^-4 M. Calculate [H_3O^+] for a solution with [OH^+] = 9.29 times 10^-9 M. Calculate [H_3O^+] for a solution of pH = 3.9. Calculate [H_3O^+] for a solution of pH = 12.72. Calculate [OH^-] for a solution of pH = 11.6. Calculate the pH of a solution with [OH^-] = 6.38 times 10^-4 M. Write the conjugate base for each of the following compounds: H_2PO_4^- NH_4^+ HSO_4^- Write the conjugate acid for each of the following compounds: SO_4^2- HCO_3^- H_2O

Explanation / Answer

1) On the pH scale the values lie between 0 and 14 with the mid point of pH = 7.0 as neutral solution. the solutions with pH< 7.0 are considered as acidic and the solutions with pH > 7.0 are basic.

a)pH = 2.1 acidic

b) pH = 7.0 neutral

c) pH = 9.2 basic

d) pH = 6.8 acidic

Q2) pH is defined as the negative logarithm (base 10) of [H+] .

Thus when [H+} = 1.0x 10-12 the pH of solution =- log 1.0x 10-12 = 12

so pH = 12

3) [H+] = 3.75 x 10 -4  

pH = - log (3.75 x 10 -4 )

= 4 - log 3.75

= 3.426

4) [OH-] = 9.29 x10-9  

Similar to pH , we can define pOH alo as

pOH = - log [OH-] and the pH of such solution is given as pH = pKw - pOH = 14 - pOH

thus pOH = - log( 9.29 x10-9 )

= 9- log 9.29 = 8.032

and pH of solution = 14 - 8.032 = 5.9680

5) Given pH = 3.9

Then [H+] = antilog of (-3.9)

= 1.26 x10-4 M

6) Given pH = 12.72

[H+] = antilog (-12.72 0

= 1.905 x10-13 M

7) Given pH = 11.6 or pOH = 14- 11.6 = 2.4

Then [OH-] = antilog (-2.4) = 3.981x 10-3 M

8) Given [OH-] = 6.38x 10-4 M

then pOH = - log (6.38x 10-4 )

= 4 - log 6.38

= 3.1952

and the pH of the solution = 14 -3.1952=10.8048

9) An accid and its conjugate base differ by a proton, an acid loses aproton (H+) to become its conjugate base.

acid conjugate base

H2PO4- HPO4-2

NH4+   NH3

HSO4-   SO4-2

10) base conjugate acid

SO4-2   HSO4-

   HCO3-   H2CO3

H2O H3O+

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