Question

What is the pH of a solution with a hydroxide concentration of 1.65×10 −5
M

Answer 1

pH = 9.217

Step-by-step explanation:

When water undergoes self dissociation, since it is an amphiprotic solvent:

`{ \rm{H _(2) O _((l)) \: \: . _( \leftarrow)^( \rightarrow) }} \: \: { \rm{H {}^( + ) _((aq)) \: \: + \: \: OH {}^( - ) _((aq)) }}`

From the dissociation constant (hydrolysis constant, Kh):

`{ \rm{k _(h) = \frac{[H {}^( + ) ][OH {}^( - ) ] }{[H _(2) O]} }} \\`

But activity of water is 1, so [H2O] = 1

`{ \rm{k _(h) = {[H {}^( + ) ][OH {}^( - ) ]}}}`

But Kh = 1 × 10^-14, and [OH-] = 1.65 × 10^-5

`\rm{(1 * {10}^( - 14) ) = {[H {}^( + ) ] * (1.65 * {10}^( - 5)) }} \\ \\ { \rm{{[H {}^( + ) ]}}}{ \rm{ = 6.06x {10}^( - 10) }}`

From the formula of pH;

`{ \rm{p{H = - log {[H {}^( + ) ]}}}} \\ { \rm{p{H }} = - log(6.06 * {10}^( - 10) ) } \\ \\ { \underline{ \rm{ \: pH = 9.217 \: }}}`