A solution of hydrochloric acid of unknown concentration was titrated with 0.16 M NaOH. If a 350-mL sample of the HCl solution required exactly 17 mL of the NaOH solution to reach the equivalence point, what was the pH of the HCl solution?

The correct answer is 2.1

Explanation:

The neutralization reaction involved in the titration is between a strong acid (HCl) and a strong base (NaOH), as follows:

HCl + NaOH → NaCl + H₂O

According to this, 1 equivalent of HCl reacts with 1 equivalent of NaOH. The equivalence point is the point at which the amount of NaOH added reacted completely with the amount of HCl.

moles of acid = moles of base

moles HCl = moles NaOH

Ca x Va = Cb x Vb

Ca = (Cb x Vb)/Va = (0.16 M x 17 mL)/(350 mL) = 7.8 x 10⁻³ M

Since HCl is a strong acid, we calculate the pH of the solution directly from the concentration of acid:

[H⁺]= Ca = 7.8 x 10⁻³ M

pH = -log [H⁺] = -log (7.8 x 10⁻³ M) = 2.11

Therefore, the pH of the HCl solution is 2.1

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.