Question

To a `1*L` volume of nitric acid solution for `[HNO_3]=0.440*mol*L^-1`, was added a mass of `17.64*g` mass of sodium hydroxide. What is `pH` of the resultant solution?

Answer 1

12

Explanation:

after the reaction it will remain 0,01 mol of NaOH in 1 L of water . since NaOH is completely dissociated you have also `[OH^-]= 10^-2 (mol)/L`while the other ions are ininfluent for calculate the pH .
Since for water `K_w= [OH^-] xx [H^+] = 10 ^(-14)`
`[H^+]= 10^(-12)`
`pH =-Log[H^+]=-Log10^(-12)=12`

Answer 2

`pH=11`

Explanation:

Consider the reaction....

`HCl(aq) + NaOH(aq) rarr NaCl(aq) + H_2O(l)`

And given this reaction, there are `(0.450-0.449)*mol` of `NaOH` dissolved in `1*L` volume of solution....

`[NaOH]-=(0.001*mol)/(1*L)=0.001*mol*L^-1`

`pOH=-log_10[HO^-]=-log_(10)10^-2=-(-2)=2`

But `pH+pOH=14`...so `pH=...?`