How many mL of 2.0M hydrochloric acid is required to neutralize 100. ML of 0.80 M potassium hydroxide?

Question

photon336 · Answer

well @ashley644 we can use this formula: 

$$M_{a}V_{a} = M_{b}V_{b}$$

the little a means acid: so it's molarity of the acid times the volume of the acid = 

the molarity of the base times the volume of the base. 

taking a deeper look at this formula, something interesting happens here. 

what this formula is saying is that the number of moles of acid is equal to the number of moles of base when we have a neutralization reaction. 

$$M_{a}V_{a} = Molarity*volume = \frac{ moles }{ L }*(L) = moles $$

First ask yourself what you are looking for:

photon336 · Answer

Be sure to convert mL to L by multiplying by dividing by 1,000

100mL/1000 = 0.1 L of 0.8 M potassium hydroxide this is our base

2.0 M HCL our acid.  

$$\frac{ M_{base}*V_{base} }{ M_{acid} } = Volume~of~acid $$

$$\frac{ 0.8M*0.1L }{ 2.0M } = Volume_{HCL}$$