# Question 6cd6a

Apr 18, 2017

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
$\text{1mole of H+"/"1mole of acid.}$ 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 $O {H}^{-}$

HCl is a strong acid thus

moles of HCl = moles of ${H}^{+}$

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

$L \cdot M = \text{moles}$

$10 \cdot {10}^{-} 3 L \cdot 0.1 M = \text{0.001mole}$

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

$\left(0.001 m o l + 0.0012 m o l\right) {H}^{+} = \text{moles of OH ions }$

$0.0022 m o l o f H + = \text{moles of OH ions}$

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

$N a O \text{H is a strong base thus}$

moles of $N a O H$ = moles of $O {H}^{-}$

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 $O {H}^{-}$

0.021mole of OH^-)# is left which means 0.021mole of $N a O H$

Now find the molarity of NaOH left

$M = \frac{\text{moles}}{L}$

Where L is 25ml = 0.025L

$M = \text{0.021mol"/"0.025L } = 0.84 M$

Now to the second part

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

$M = \text{0.022mol"/"0.025L} = 0.88 M$