Question

Plz answer correctly....The concentration of either the H+ ion or the OH − ion is given for four aqueous solutions at 298 K.?

Answer 1

Explanation :

Solution for part (a) :

Concentration of $H^+$ = $8.0\times 10^{-13}M$

As we know that,

$[H^+][OH^-]=10^{-14}$

Now put the value of concentration of hydrogen ion, we get concentration of hydroxide ion.

$(8.0\times 10^{-13})\times [OH^-]=10^{-14}$

$[OH^-]=1.25\times 10^{-2}M$

From this we conclude that, the concentration of hydroxide ion is more than the hydrogen ion that means the solution is basic in nature.

Solution for part (b) :

Concentration of $OH^-$ = $4.0\times 10^{-7}M$

As we know that,

$[H^+][OH^-]=10^{-14}$

Now put the value of concentration of hydroxide ion, we get concentration of hydrogen ion.

$[H^+]\times (4.0\times 10^{-7})=10^{-14}$

$[H^+]=0.25\times 10^{-7}M$

From this we conclude that, the concentration of hydrogen ion is more than the hydroxide ion that means the solution is acidic in nature.

Solution for part (c) :

Concentration of $OH^-$ = $5.0\times 10^{-3}M$

As we know that,

$[H^+][OH^-]=10^{-14}$

Now put the value of concentration of hydroxide ion, we get concentration of hydrogen ion.

$[H^+]\times (3.0\times 10^{-3})=10^{-14}$

$[H^+]=0.25\times 10^{-7}M$

From this we conclude that, the concentration of hydroxide ion is more than the hydrogen ion that means the solution is basic in nature.

Solution for part (d) :

Concentration of $H^+$ = $8.0\times 10^{-5}M$

As we know that,

$[H^+][OH^-]=10^{-14}$

Now put the value of concentration of hydrogen ion, we get concentration of hydroxide ion.

$(8.0\times 10^{-5})\times [OH^-]=10^{-14}$

$[OH^-]=1.25\times 10^{-2}M$

From this we conclude that, the concentration of hydrogen ion is more than the hydroxide ion that means the solution is acidic in nature.

Answer 2

Definitions needed:  

pH = -log[H^+]  

pOH = -log[OH^-]  

pH + pOH = 14  

[H^+]*[OH^-] = 10^-14  

(a) [H+] = 3.0 10-13 M  

[OH ‾ ] = _____ M  

Acidic, Basic, or Neutral?  

[H^+]*[OH^-] = 10^-14  

[OH^-] = (10^-14)/[H^+]  

[OH^-] = (10^-14)/(3.0 10-13 M) = 0.033 M  

That's a high concentration of [OH^-], so the solution is basic  

(b) [OH ‾ ] = 3.0 10-7 M  

[H+] = ______M  

Acidic, Basic, or Neutral?  

[H^+]*[OH^-] = 10^-14  

[H^+]* = (10^-14)/[OH^-]  

[H^+]* = (10^-14)/(3.0 10-7 M) = 3.33 x 10^-8  

That would be a slightly basic solution (pH just higher than 7).