Question

How much 0.170 M KOH is required to completely neutralize 50.0 mL of 0.170 M HCIO;?

Answer 1

The first step to solve this question is to state the neutralization reaction between KOH and HClO:

`KOH+HClO\rightarrow KClO+H_2O`

Now, find the amount of moles of HClO present in 50mL of 0.170M HClO:

`50.0mL\cdot(1L)/(1000mL)\cdot(0.170mol)/(L)=0.0085mol`

According to the equation, 1 mole of KOH reacts with 1 mole of HClO. Use this ratio to find the amount of moles of KOH that react with 0.0085moles of HClO:

`0.0085molHClO\cdot(1molKOH)/(1molHClO)=0.0085molKOH`

Multiply this amount of moles of KOH by the inverse of its conversation:

`0.0085molKOH\cdot(1L)/(0.170molKOH)=0.05L`

It means that 0.05L of 0.170M KOH are needed to neutralize 50mL of 0.170M HClO.