Answer:
0.190 M
Explanation:
Let's consider the neutralization reaction between HCl and NaOH.
HCl + NaOH = NaCl + H2O
11.9 mL of 0.160 M NaOH were used. The reacting moles of NaOH were:
0.0119 L × 0.160 mol/L = 1.90 × 10⁻³ mol
The molar ratio of HCl to NaOH is 1:1. The reacting moles of HCl are 1.90 × 10⁻³ moles.
1.90 × 10⁻³ moles of HCl are in 10.0 mL of solution. The molarity of HCl is:
M = 1.90 × 10⁻³ mol / 10.0 × 10⁻³ L = 0.190 M
Answer:
The initial concentration of HCl was 0.1904 M
Explanation:
Step 1: Data given
Volume of HCl solution = 10.0 mL = 0.010 L
Volume of a NaOH solution = 11.9 mL = 0.0119 L
Molarity of NaOH solution = 0.160 M
Step 2: The balanced equation
HCl + NaOH → NaCl + H2O
Step 3: Calculate the concentration of HCl
C1*V1 = C2*V2
⇒with C1 = the concentration HCl = TO BE DETERMINED
⇒with V1 = the volume of HCl = 0.010 L
⇒with C2 = the concentration of NaOH = 0.160 M
⇒with V2 = the volume of NaOH = 0.0119 L
C1 * 0.010 L = 0.160 M * 0.0119 L
C1 = (0.160 M * 0.0119 L) / 0.010 L
C1 = 0.1904 M
The initial concentration of HCl was 0.1904 M
