Question

Question #6cd6a

Answer 1

For neutralization

mol of H+ ions of acid = mol of OH- ions of base

So first you need to know the H+ of 10ml of 0.1 M HCl

The number `H^+` ions in the solution is the concentration of HCl because strong acids dissociate completely thus producing
`"1mole of H+"/"1mole of acid."` This is only applicable to strong acids. The same is the case with strong base but instead of `H^+` they produce strong conjugate bases like `OH^(-)`

HCl is a strong acid thus

moles of HCl = moles of `H^+`

Calculate no. of moles in 10mL of 0.1M HCl

`L * M = "moles"`

`10* 10^-3L * 0.1M = "0.001mole"`

0.001mole of H+ couldn't neutralize/neutralise 25ml of NaOH of a certain concentration

8ml of 0.15M of HCl was further needed

Calculate the moles of H+ using the same way

= 0.0012mol of H+

So this means

0.001 mol of H+ needs more H+ ions to neutralize/neutralise the base and it it needs 0.0012mol of H+ again. it means

`(0.001mol + 0.0012mol)H^+ = "moles of OH ions "`

`0.0022mol of H+ = "moles of OH ions"`

Thus there are 0.0022mol of OH- in the NaOH solution

`NaO"H is a strong base thus"`

moles of `NaOH` = moles of `OH^-`

Thus 0.022mol of NaOH is present in 25ml solution of NaOH

Now we know the initial amout of NaOH and HCl

As this is titrating of a strong acid with a strong base the addition of the base results in decrease in `H^+` ions. 1 mol of OH- results in decrease of 1 mol of H+ and viceversa.

0.022mol of OH - 0.001mole of H = 0.021mole of `OH^-`

0.021mole of `OH^-)` is left which means 0.021mole of `NaOH`

Now find the molarity of NaOH left

`M = "moles"/L`

Where L is 25ml = 0.025L

`M = "0.021mol"/"0.025L " = 0.84M`

Now to the second part

Since we know the amount of NaOH in moles and litres of water dissolving it find the molarity

`M = "0.022mol"/"0.025L" = 0.88M`