Whay volume of a 0.500 M HCl solution is needed to neutralize each: 14.0 ml of a 0.300 M NaOH and 19.0ml of a 0.200 M Ba(OH)2

Question

anonymous · Answer

For HCl being neutralized in the first case with NaOH , you first need an equation  we have 

$$HCl + NaOH \rightarrow NaCl+ H _{2}O $$ both sides are balanced, so we can now find how many moles we have of NaOH and use the ratio of moles of HCl to NaOH to find the moles of HCl . 

As you know Molarity $$M = moles \div liters(of solution) $$ 
we can rearrange this equation to find the moles of NaOH so it wold be 
$$moles = Molarity \times liters(volume)  $$
we know molarity of NaOH to be 0.300 and the volume is 14.0ml which we have to convert to L so 0.014L .

The values of moles we get is $$4.3\times10^{-3}  $$ which is out moles of NaOH then apply the molar ratio which is 1:1 for NaOH and HCl giving you the same amount of moles of HCl . So now with moles of HCl we apply the equation except we look for Liters(Volume ) 
$$Volume = moles \div MolarityofHCL $$
which gives us 8.4x10-3 L of HCl needed . In ml it's 8.4mL 

You can use this same method to find the value of HCl needed for Ba(OH)2 which comes out to be 0.0152L or 15.2ml

anonymous · Answer

Oh wow! Thank you very much. I made it half way until i got stuck.

anonymous · Answer

No problem, these questions are pretty easy once you break them down a go step by step :)