calculate the ph of each solution at 25 °c.\n1.0×10⁻⁵ m hcl ph = \n0.01 m hno₃ ph = \n1.0×10⁻⁵ m naoh ph =…

Explanation:

Step1: Recall pH formula for acids

For strong acids, $[H^+]$ is equal to the acid concentration. The pH formula is $pH = -\log[H^+]$.

Step2: Calculate pH of $1.0\times 10^{-5}\ M\ HCl$

$HCl$ is a strong - acid, so $[H^+]=1.0\times 10^{-5}\ M$. Then $pH = -\log(1.0\times 10^{-5}) = 5$.

Step3: Calculate pH of $0.01\ M\ HNO_3$

$HNO_3$ is a strong - acid, so $[H^+]=0.01\ M = 1.0\times 10^{-2}\ M$. Then $pH=-\log(1.0\times 10^{-2}) = 2$.

Step4: Recall relationship for bases

For strong bases, first find $[OH^-]$. Then use $K_w=[H^+][OH^-]=1.0\times 10^{-14}$ at $25^{\circ}C$ to find $[H^+]$, and then $pH = -\log[H^+]$.

Step5: Calculate pH of $1.0\times 10^{-5}\ M\ NaOH$

$NaOH$ is a strong - base, so $[OH^-]=1.0\times 10^{-5}\ M$. Then $[H^+]=\frac{K_w}{[OH^-]}=\frac{1.0\times 10^{-14}}{1.0\times 10^{-5}} = 1.0\times 10^{-9}\ M$. So $pH = -\log(1.0\times 10^{-9})=9$.

Step6: Calculate pH of $0.1\ M\ KOH$

$KOH$ is a strong - base, so $[OH^-]=0.1\ M = 1.0\times 10^{-1}\ M$. Then $[H^+]=\frac{K_w}{[OH^-]}=\frac{1.0\times 10^{-14}}{1.0\times 10^{-1}} = 1.0\times 10^{-13}\ M$. So $pH = -\log(1.0\times 10^{-13}) = 13$.

Answer:

$1.0\times 10^{-5}\ M\ HCl$: $5$ $0.01\ M\ HNO_3$: $2$ $1.0\times 10^{-5}\ M\ NaOH$: $9$ $0.1\ M\ KOH$: $13$