chemistry homework\nchapter 19: acids and bases\nshow all work for full credit!\nname _efrain hernandez_\n1…
chemistry homework\nchapter 19: acids and bases\nshow all work for full credit!\nname _efrain hernandez_\n1. given the concentration of either hydrogen ion or hydroxide ion, use the ion product constant for water to calculate the concentration of the other ion at 298k.\n a. h+=1.0×10 - 4 m\n b. oh-=1.3×10 - 2 m\n2. calculate the ph at 298k of solutions having the following ion concentrations.\n a. h+=1.0×10 - 4 m\n b. h+=5.8×10 - 11 m\n3. calculate the poh and ph at 298k of solutions having the following ion concentrations.\n a. oh-=1.0×10 - 12 m\n b. oh-=1.3×10 - 2 m\n4. calculate the ph of each of the following strong acid or strong base solutions at 298k.\n a. 2.6×10 - 2 m hcl\n b. 0.28 m hno3\n c. 7.5×10 - 3 m naoh\n d. 0.44 m koh
Answer
1.
a.
Explanation:
Step1: Recall ion - product constant formula
The ion - product constant of water $K_w=[H^+][OH^-]=1.0\times10^{- 14}$ at 298K. We know $[H^+]=1.0\times10^{-4}\ M$, and we want to find $[OH^-]$. [ [OH^-]=\frac{K_w}{[H^+]} ]
Step2: Substitute values
Substitute $K_w = 1.0\times10^{-14}$ and $[H^+]=1.0\times10^{-4}\ M$ into the formula. [ [OH^-]=\frac{1.0\times10^{-14}}{1.0\times10^{-4}} = 1.0\times10^{-10}\ M ]
b.
Step1: Recall ion - product constant formula
Using $K_w=[H^+][OH^-]=1.0\times10^{-14}$ at 298K. We know $[OH^-]=1.3\times10^{-2}\ M$, and we want to find $[H^+]$. [ [H^+]=\frac{K_w}{[OH^-]} ]
Step2: Substitute values
Substitute $K_w = 1.0\times10^{-14}$ and $[OH^-]=1.3\times10^{-2}\ M$ into the formula. [ [H^+]=\frac{1.0\times10^{-14}}{1.3\times10^{-2}}\approx7.7\times10^{-13}\ M ]
2.
a.
Explanation:
Step1: Recall pH formula
The pH is defined as $pH =-\log[H^+]$. Given $[H^+]=1.0\times10^{-4}\ M$. [ pH=-\log(1.0\times10^{-4}) ]
Step2: Calculate pH
Since $\log(1.0\times10^{-4})=- 4$, then $pH = 4$.
b.
Step1: Recall pH formula
Using $pH =-\log[H^+]$, with $[H^+]=5.8\times10^{-11}\ M$. [ pH=-\log(5.8\times10^{-11}) ]
Step2: Calculate pH
$pH=-( \log5.8+\log(10^{-11}))=-(0.76 - 11)=10.24$
3.
a.
Explanation:
Step1: Calculate pOH
The pOH is defined as $pOH=-\log[OH^-]$. Given $[OH^-]=1.0\times10^{-12}\ M$. [ pOH=-\log(1.0\times10^{-12}) = 12 ]
Step2: Calculate pH
Since $pH + pOH=14$ at 298K, then $pH=14 - pOH$. Substitute $pOH = 12$ into the formula, so $pH = 2$.
b.
Step1: Calculate pOH
Using $pOH=-\log[OH^-]$, with $[OH^-]=1.3\times10^{-2}\ M$. [ pOH=-\log(1.3\times10^{-2})=-( \log1.3+\log(10^{-2}))=-(0.11-2)=1.89 ]
Step2: Calculate pH
Since $pH + pOH = 14$ at 298K, then $pH=14 - pOH$. Substitute $pOH = 1.89$ into the formula, so $pH=12.11$
4.
a.
Explanation:
Step1: Determine $[H^+]$
HCl is a strong acid, so $[H^+]=2.6\times10^{-2}\ M$.
Step2: Calculate pH
Using $pH =-\log[H^+]$, then $pH=-\log(2.6\times10^{-2})=-( \log2.6+\log(10^{-2}))=-(0.41-2)=1.59$
b.
Step1: Determine $[H^+]$
$HNO_3$ is a strong acid, so $[H^+]=0.28\ M$.
Step2: Calculate pH
Using $pH =-\log[H^+]$, then $pH=-\log(0.28)=-( \log2.8 - 1)=0.55$
c.
Step1: Determine $[OH^-]$
NaOH is a strong base, so $[OH^-]=7.5\times10^{-3}\ M$.
Step2: Calculate pOH
Using $pOH=-\log[OH^-]$, then $pOH=-\log(7.5\times10^{-3})=-( \log7.5+\log(10^{-3}))=-(0.88 - 3)=2.12$
Step3: Calculate pH
Since $pH + pOH=14$ at 298K, then $pH=14 - pOH=11.88$
d.
Step1: Determine $[OH^-]$
KOH is a strong base, so $[OH^-]=0.44\ M$.
Step2: Calculate pOH
Using $pOH=-\log[OH^-]$, then $pOH=-\log(0.44)=-( \log4.4 - 1)=0.36$
Step3: Calculate pH
Since $pH + pOH = 14$ at 298K, then $pH=14 - pOH = 13.64$
Answer:
- a. $[OH^-]=1.0\times10^{-10}\ M$; b. $[H^+]\approx7.7\times10^{-13}\ M$
- a. $pH = 4$; b. $pH=10.24$
- a. $pOH = 12$, $pH = 2$; b. $pOH=1.89$, $pH=12.11$
- a. $pH=1.59$; b. $pH = 0.55$; c. $pH=11.88$; d. $pH=13.64$