Answer:
0.36 M
Explanation:
Let's consider the following neutralization reaction.
HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)
25 mL of 0.83 M NaOH reacted. The moles of NaOH that reacted are:
0.025 L × 0.83 mol/L =0.021 mol
The molar ratio of NaOH to HCl is 1:1. The moles of HCl that reacted are 0.021 moles.
0.021 moles are contained in 58 mL of HCl. The molar concentration of HCl is:
M = 0.021 mol/0.058 L
M = 0.36 M
Answer: 0.36 M
Explanation:
25 mL NaOH × 1L/1000 mL × 0.83 mol NaOH/ 1L × 1 mol HCl/ 1mol NaOH =2.075×10^−2 mol HCl
[HCl]= (2.075 × 10^−2mol / (58mL×1 L/ 1000mL)) = 0.358 M --> 0.36 M
