Free Printable pH Worksheets - Free Printable

Here are the step-by-step calculations to solve the problems on the worksheet.

Problem 1: Complete the following table

We need to use the following relationships to fill in the blanks:
1. $pH = -\log[H^+]$
2. $pOH = -\log[OH^-]$
3. $pH + pOH = 14$
4. $[H^+] \times [OH^-] = 1.0 \times 10^{-14}$

Row 1: Given $pH = 3.5$
* $[H^+]$: $10^{-3.5} \approx 3.16 \times 10^{-4}$ M
* pOH: $14 - 3.5 = 10.5$
* $[OH^-]$: $10^{-10.5} \approx 3.16 \times 10^{-11}$ M

Row 2: Given $[H^+] = 5.8 \times 10^{-7}$
* pH: $-\log(5.8 \times 10^{-7}) \approx 6.24$
* pOH: $14 - 6.24 = 7.76$
* $[OH^-]$: $10^{-7.76} \approx 1.74 \times 10^{-8}$ M

Row 3: Given $[OH^-] = 4.2 \times 10^{-2}$
* pOH: $-\log(4.2 \times 10^{-2}) \approx 1.38$
* pH: $14 - 1.38 = 12.62$
* $[H^+]$: $10^{-12.62} \approx 2.38 \times 10^{-13}$ M

Row 4: Given $pOH = 8.2$
* pH: $14 - 8.2 = 5.8$
* $[H^+]$: $10^{-5.8} \approx 1.58 \times 10^{-6}$ M
* $[OH^-]$: $10^{-8.2} \approx 6.31 \times 10^{-9}$ M

Row 5: Given $[H^+] = 4.2 \times 10^{-5}$
* pH: $-\log(4.2 \times 10^{-5}) \approx 4.38$
* pOH: $14 - 4.38 = 9.62$
* $[OH^-]$: $10^{-9.62} \approx 2.40 \times 10^{-10}$ M

Row 6: Given $pOH = 2.4$
* pH: $14 - 2.4 = 11.6$
* $[H^+]$: $10^{-11.6} \approx 2.51 \times 10^{-12}$ M
* $[OH^-]$: $10^{-2.4} \approx 3.98 \times 10^{-3}$ M

Row 7: Given $pH = 10.1$
* pOH: $14 - 10.1 = 3.9$
* $[H^+]$: $10^{-10.1} \approx 7.94 \times 10^{-11}$ M
* $[OH^-]$: $10^{-3.9} \approx 1.26 \times 10^{-4}$ M

Row 8: Given $[OH^-] = 7.2 \times 10^{-3}$
* pOH: $-\log(7.2 \times 10^{-3}) \approx 2.14$
* pH: $14 - 2.14 = 11.86$
* $[H^+]$: $10^{-11.86} \approx 1.38 \times 10^{-12}$ M

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Problem 2: Calculate the values of both pH and pOH

Row 1: $0.020$ M HCl
* HCl is a strong acid, so $[H^+] = 0.020$ M.
* pH: $-\log(0.020) \approx 1.70$
* pOH: $14 - 1.70 = 12.30$

Row 2: $0.0050$ M NaOH
* NaOH is a strong base, so $[OH^-] = 0.0050$ M.
* pOH: $-\log(0.0050) \approx 2.30$
* pH: $14 - 2.30 = 11.70$

Row 3: Blood sample $7.2 \times 10^{-8}$ M of $H^+$
* pH: $-\log(7.2 \times 10^{-8}) \approx 7.14$
* pOH: $14 - 7.14 = 6.86$

Row 4: $0.00035$ M KOH
* KOH is a strong base, so $[OH^-] = 0.00035$ M.
* pOH: $-\log(0.00035) \approx 3.46$
* pH: $14 - 3.46 = 10.54$

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Problem 3: What is the $[H_3O^+]$ in $200.0$ mL of $0.0010$ M KOH?

1. KOH is a strong base. The concentration of hydroxide ions $[OH^-]$ is equal to the concentration of the solution: $0.0010$ M.
* *Note: The volume ($200.0$ mL) is extra information and is not needed to find the concentration.*
2. Use the water constant ($K_w$) to find the hydronium concentration:
$$[H_3O^+] \times [OH^-] = 1.0 \times 10^{-14}$$
$$[H_3O^+] \times 0.0010 = 1.0 \times 10^{-14}$$
3. Solve for $[H_3O^+]$:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{0.0010}$$
$$[H_3O^+] = 1.0 \times 10^{-11} \text{ M}$$

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Problem 4: Tomato juice has a pH of 4.20. Calculate the $[H_3O^+]$ and $[OH^-]$ in tomato juice.

1. Calculate $[H_3O^+]$:
$$[H_3O^+] = 10^{-pH}$$
$$[H_3O^+] = 10^{-4.20}$$
$$[H_3O^+] \approx 6.31 \times 10^{-5} \text{ M}$$

2. Calculate $[OH^-]$:
First, find the pOH:
$$pOH = 14 - pH$$
$$pOH = 14 - 4.20 = 9.80$$
Now, find $[OH^-]$:
$$[OH^-] = 10^{-pOH}$$
$$[OH^-] = 10^{-9.80}$$
$$[OH^-] \approx 1.58 \times 10^{-10} \text{ M}$$

Final Answer:
1. Table Solutions:
* Row 1: $[H^+] = 3.2 \times 10^{-4}$, $[OH^-] = 3.2 \times 10^{-11}$, pOH = 10.5
* Row 2: pH = 6.24, $[OH^-] = 1.7 \times 10^{-8}$, pOH = 7.76
* Row 3: pH = 12.62, $[H^+] = 2.4 \times 10^{-13}$, pOH = 1.38
* Row 4: pH = 5.8, $[H^+] = 1.6 \times 10^{-6}$, $[OH^-] = 6.3 \times 10^{-9}$
* Row 5: pH = 4.38, $[OH^-] = 2.4 \times 10^{-10}$, pOH = 9.62
* Row 6: pH = 11.6, $[H^+] = 2.5 \times 10^{-12}$, $[OH^-] = 4.0 \times 10^{-3}$
* Row 7: $[H^+] = 7.9 \times 10^{-11}$, $[OH^-] = 1.3 \times 10^{-4}$, pOH = 3.9
* Row 8: pH = 11.86, $[H^+] = 1.4 \times 10^{-12}$, pOH = 2.14

2. pH and pOH Calculations:
* 0.020 M HCl: pH = 1.70, pOH = 12.30
* 0.0050 M NaOH: pH = 11.70, pOH = 2.30
* Blood Sample: pH = 7.14, pOH = 6.86
* 0.00035 M KOH: pH = 10.54, pOH = 3.46

3. $[H_3O^+]$ in KOH solution:
$1.0 \times 10^{-11}$ M

4. Tomato Juice Concentrations:
$[H_3O^+] = 6.3 \times 10^{-5}$ M
$[OH^-] = 1.6 \times 10^{-10}$ M