Question

Calculate the resulting ph if 365 ml of 2.88 m hno3 is mixed with 335 ml of 1.10 m ca(oh)2 solution. be aware of stoichiometry!

Answer 1

0.350 is the answer.

Answer 2

Answer: pH is 0.34

Explanation: Moles can be calculated by using the formula:

`\text{no of moles of } HNO_3}={\text{Molarity}* {\text{Volume in L}}`

`\text{no of moles}={2.88M}* {0.365L}=1.05moles`

`HNO_3\rightarrow H^++NO_3^_`

Moles of
`H^+` ions = 1.05

`\text{no of moles of } Ca(OH)_2}={\text{Molarity}* {\text{Volume in L}}`

`\text{no of moles}={1.10M}* {0.335L}=0.36`

`Ca(OH)_2\rightarrow Ca^(2+)+2OH^-`

Moles of
`OH^-` ions =
`2* 0.36=0.73moles`

Thus 0.73 moles of
`Ca(OH)_2` will neutralize 0.73 moles of
`HNO_3` and (1.05-0.73)=0.32 moles of
`HNO_3` will be left.

Molarity of
`H^+=(moles)/(total volume in L)=(0.32moles)/(0.7L)=0.46M`

`pH=-\log[H^+]=-\log[0.46]=0.34`