Question

1. complete the reaction hcl + h₂o → h₃o⁺¹ + 1

there is complete ionization

1. what is the ph of a 0.001000 m hcl solution ? ph = -log( 2 ) = 3

a. acetic b. hydrochloric c. phosphoric d. perchloric e. hydrofluoric
f. sulfuric g. sulfurous h. hydrosulfuric i. oh⁻¹ j. h₃o⁺¹ k. so₄⁻²
l. mg⁺² m. cl⁻¹ n. hpo₃⁻² o. h₂po₃⁻¹ p. co₃⁻² q. hco₃⁻¹
r. h₂s s. hs⁻¹ t. s⁻² u. 0.2500 v. 0.1000 w. 0.02500
x. 1.000 × 10⁻³ y. 3.0000 z. 1 × 10⁻¹⁴ aa. 2.6990 bb. 1.000 × 10⁻¹¹
cc. 11.0000 dd. 2 ee. 1 ff. 2.000 × 10⁻³ gg. 5.000 × 10⁻¹²
hh. 11.3010 ii. 0.01250 jj. 0.22100 kk. 0.05656 ll. acid
mm. conjugate acid oo. base pp. conjugate base

Explanation:

Step1: Complete the reaction

For the reaction \(HCl + H_2O
ightarrow H_3O^{+1}+\)___, since \(HCl\) ionizes completely in water, the other ion formed is \(Cl^{-1}\) (M).

Step2: Calculate the \(pH\)

For a \(0.001000\ M\ HCl\) solution, \(HCl\) is a strong acid and completely ionizes, so \([H_3O^{+}]=0.001000\ M = 1.000\times 10^{-3}\ M\) (X).
Using the formula \(pH =-\log[H_3O^{+}]\), we substitute \([H_3O^{+}]=1.000\times 10^{-3}\) into the formula: \(pH=-\log(1.000\times 10^{-3})\).
Since \(\log(1.000\times 10^{-3})=- 3\), then \(pH = 3.0000\) (Y).

Answer:

1. M. \(Cl^{-1}\)
2. X. \(1.000\times 10^{-3}\), Y. \(3.0000\)