Question

30 ml of 0.10M NaOH neutralized 25.0ml of HCI
Determine the concentration of the HCI?

1. Write the balanced chemical equation for the reaction
NaOH + HCI --> NaCI + H20

2. Extract the relevant information from the qustion:
NaOH v= 30mL, M=0.10

HCI v= 25.0 mL, M=?

3. Convert to Liters
NaOH v= 0.03 L, M= 0.10M

HCI v=0.025L, M=?

4. Calculate moles NaOh:
n(NaOH) = M x V = 0.10 moles/L x 0.03L = 3x10^-3 moles

5. From the balanced chemical equation find the mole ratio
NaOH:HCI = 1:1

6. Find the number of moles of HCI that were titrated
NaOH: HCI is 1:1;
so n(NaOH) = n(HCI) = 0.003 moles at neutralization

7. Calculate the concentration of HCI:
M=n divided by V n=0.003 mol, V=0.025L

M(hci)= 0.003moles divided by 0.025L = 0.12 moles/L or 0.12

Answer 1

@Vocaloid @Shadow Might be able to help

Answer 2

for this rxn moles NaOH = moles HCl

so moles HCl = moles NaOH = molarity * volume *** make sure to convert volume to L first ***

then M HCl = moles HCL/volume HCL, again making sure to convert to L

Answer 3

jeez im not good at this

Answer 4

lol

Answer 5

can u help me worth through it @Vocaloid