Chemistry Acids/Bases Questions!

Hi, a few Chem problems I just can't get. Any help is greatly appreciated!

1. Calculate the molarity of an HCl solution if 30.8 mL of the solution is neutralized by 15.5 mL of 0.451 M NaOH.

2. The net ionic equation for the neutralization reaction between solutions of potassium hydroxide and hydrochloric acid is which of the following?

A. H+(aq) + OH -(aq) --> H2O(l)
B.K+(aq) + OH -(aq) + Cl -(aq) --> KCl(aq) + H2O(l)
C. K+(aq) + Cl -(aq) --> KCl(aq)
D. KOH(aq) + HCl(aq) --> H2O(l) + KCl(aq)

1.
Both HCl and NaOH are both dibasic. This meant that they will react in a 1:1 ratio.
HCl + NaOH -> NaCl + H2O

we will first find the number of moles of NaOH used in this reaction.
15.5mL = 0.0155 dm^3

Number of moles of NaOH
= volume of NaOH used * molarity of NaOH used
=0.0155 * 0.451 = 0.00699mol

Number of moles of HCl = Number of moles of NaOH (1:1 ratio)

Volume of HCl used = 0.0308dm^3

Molarity = Number of moles / Volume in dm^3
=0.00699 / 0.0308
=0.227 mol/dm^3 (3 significant figures)

2. A

1) Volume of HCl which is a strong acid = 30.8ml=0.0308L
You are told that 15.5ml of 0.451M of the strong base NaOH to neutralize the acid.
NaOH + HCl = NaCl + H2O
You can see that the ratio is 1:1
number of moles of base = 0.0155*0.451=C*V=6.99*10^-3mol=number of moles of HCl
C=n/v ( Concentration of acid = Molarity )
C=6.99*10^-3/0.0308=0.22M

2) Pottasium Hydroxide = KOH
Hydrochloric acid = HCl
H+ will react with OH- to give you water to make it more neutral
Your answer is A

The answer to the 2nd question should be A