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Answer:

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Explanation:

9.) 10-2.76 =0.0174 [H30+]= 1.74*10-3 M

10.)10-3.65=0.00224  [H3O+] =2.24*10-2 M

11.)10-3.65=0.00224 [OH-]= 2.224*10-4M

12.)10-6.87=0.00000135  [OH-]= 1.35*10-7M

pstnonsonjoku

We can calculate the concentration of oxonium ions  [H3O+] from the pH and the concentration of hydroxide ions [OH-]  from the pOH.

To obtain the [H3O+] of a solution of pH 2.76:

pH = - log  [H3O+]

Hence;

[H3O+] = Antilog (-pH)

[H3O+] = Antilog (-2.76)

[H3O+] = 1.74 * 10^-3 M

To obtain the [H3O+] of a solution of pH 3.65

pH = - log  [H3O+]

Hence;

[H3O+] = Antilog (-pH)

[H3O+] = Antilog (-3.65)

[H3O+] = 2.23 * 10^-4 M

To obtain the [OH-] of a solution of pOH 3.65

pOH= - log  [OH-]

Hence;

[OH-]= Antilog (-pOH)

[OH-] = Antilog (-3.65)

[OH-] = 2.23 * 10^-4 M

To obtain the [OH-] of a solution of pOH 6.87​

pOH= - log  [OH-]

Hence;

[OH-]= Antilog (-pOH)

[OH-] = Antilog (-6.87​)

[OH-] = 1.35 * 10^-7 M

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