Question

The pOH of a basic solution is 5.47. What is [OH-]? 0 of 1 point earned 2 attempts remaining The pH of a basic solution is 9.91. What is POH? 9 O of 1 point earned 2 attempts remaining The pOH of a basic solution is 2.79. What is [H-1? 10 L 0 of 1 point earned 2 attempts remaining The pH of a solution is 3.42 What is the OH solution? concentration in the 11 0 of 1 point earned artemis rema

Answer 1

8.

Answer :-  
[OH-] = 3.39 x 10-6(M)

Explanation :-

Given :- pOH = 5.47

we know that,
pOH = -log(OH

i.e.
[OH-] = antilog-poH

therefore,   

[OH-] = antilog(-5.47)

i.e.   
[OH-] = 3.39 x 10-6(M)

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9.

Answer :-  
0 = 4.09

Explanation :-

Given :- pH = 9.91

we know that,   
+poll = 14

or
pOH = 14 -

therefore,
OH = 14 -9.91 = 4.09

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10.

Answer :-
(H+) = 6.17 x 10-12(M)

Explanation :-

Given :- pOH = 2.79

we know that,
+poll = 14

​​​​​​therefore, we have,  

pH +2.79 = 14

i.e.
pH = 14 - 2.79 = 11.21

Now we know that,

(H+) = antilog(-p)

therefore,
[+] = antilog(-11.21)

i.e.   
(H+) = 6.17 x 10-12(M)

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11.

Answer :-

Explanation :-

Given :- pH = 3.42

we know that,
+poll = 14

therefore,
3.42 + p 1 = 14

i.e.   
OH = 14 - 3.42 = 10.58

Now we know that,

  
[OH-] = antilog-poH

i.e.   
[OH-] = antilog(-10.58)

i.e.
[OH-] = 2.63 x 10-11(M)

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13.

Answer :-
(WV).-01 X 92 2 = l+H

Explanation :-

Given :- pOH = 9.89

we know that
+poll = 14

therefore,
" +9.89 = 14

i.e.
pH = 14 - 9.89 = 4.11

Now, we know that,  

  
(H+) = antilog(-p)

therefore,
(H+) = antilog(-4.11)

i.e.
(WV).-01 X 92 2 = l+H

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14.

Answer :-
pOH = 12.46

Explanation :-

Given :- Molarity of HCl = 0.0290 M

we know that, HCl is strong acid snd hence it dissociate completely as  

HCl(aq) + Hg) + Cliad

Thus, we have, [H+] = 0.0290 M

we know that
pH = -log[H+

therefore,   
pH = -log[0.0290

i.e.   
P = -(-1.54)

i.e.
W = 1.54

Now we know that
+poll = 14

therefore,
1.54 +pOH = 14

i.e.
pOH = 14 - 1.54 = 12.46

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15.

Answer :-  

Explanation :-

Given :- Molarity of CsOH =
4.3 x 10-4(M)

we know that CsOH is strong base and hence it dissociate completely as

CsOH(aq) → Cstag + oHaq)

thus we have [OH-] =
4.3 x 10-4(M)

we know that

  
pOH = -log(OH

therefore,  
pOH = -log[4.3 x 10-4

i.e.   
pOH = -(-3.37) = 3.37

Now we know that  
+poll = 14

i.e.   
pH + 3.37 = 14

i.e.  
pH = 14 -3.37 = 10.63

Again we know that,
(H+) = antilog(-p)

i.e.   
(+1 = antilog(-10.63)

i.e
(H+) = 2.34 x 10-11(M)

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