Question

question 18
the ph scale for acidity is defined by $\text{ph} = -\log_{10}\text{h}^+$ where $\text{h}^+$ is the concentration of hydrogen ions measured in moles per liter (m).
a solution has a ph of 5.25.
calculate the concentration of hydrogen ions in moles per liter (m). write the answer in scientific notation with three significant digits.
the concentration of hydrogen ions is \boxed{} moles per liter.
remember to enter scientific notation: $1\times10^{-14}$ would be entered as $1^{\ast}10^{\wedge}-14$
question help: written example
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Explanation:

Step1: Start with the pH formula

The formula for pH is \( \text{pH} = -\log_{10}[\text{H}^+] \). We know the pH is 5.25, so we can rewrite the formula to solve for \( [\text{H}^+] \). First, multiply both sides by -1: \( -\text{pH} = \log_{10}[\text{H}^+] \). Then, convert from logarithmic form to exponential form. Since \( \log_{10}x = y \) is equivalent to \( x = 10^y \), we have \( [\text{H}^+] = 10^{-\text{pH}} \).

Step2: Substitute the pH value

Substitute \( \text{pH} = 5.25 \) into the formula: \( [\text{H}^+] = 10^{-5.25} \).

Step3: Calculate \( 10^{-5.25} \)

We can rewrite \( -5.25 \) as \( -5 - 0.25 \), so \( 10^{-5.25} = 10^{-5 - 0.25} = 10^{-5} \times 10^{-0.25} \). Calculate \( 10^{-0.25} \). We know that \( 10^{-0.25} = \frac{1}{10^{0.25}} \). \( 10^{0.25} = 10^{\frac{1}{4}} \approx 1.778 \) (since \( 1.778^4 \approx 10 \)). So \( 10^{-0.25} \approx \frac{1}{1.778} \approx 0.5623 \). Then, \( 10^{-5} \times 0.5623 = 5.623 \times 10^{-6} \) (because \( 0.5623 \times 10^{-5} = 5.623 \times 10^{-6} \)). Rounding to three significant digits, we get \( 5.62 \times 10^{-6} \)? Wait, no, wait. Wait, let's recalculate \( 10^{-5.25} \) more accurately. Using a calculator, \( 10^{-5.25} = e^{-5.25 \ln 10} \approx e^{-5.25 \times 2.302585093} \approx e^{-12.0885715} \)? No, that's not right. Wait, no, \( \log_{10}(x) = -5.25 \) means \( x = 10^{-5.25} \). Let's use the calculator for \( 10^{-5.25} \). \( 10^{-5.25} = 10^{-(5 + 0.25)} = 10^{-5} \times 10^{-0.25} \). \( 10^{-0.25} \) is the same as \( 10^{3/4} \) reciprocal? Wait, no, \( 0.25 = 1/4 \), so \( 10^{-1/4} \). Let's calculate \( 10^{0.25} \): \( 10^{0.25} = \sqrt[4]{10} \approx 1.77827941 \), so \( 10^{-0.25} \approx 1/1.77827941 \approx 0.562341325 \). Then, \( 10^{-5} \times 0.562341325 = 0.562341325 \times 10^{-5} = 5.62341325 \times 10^{-6} \). Rounding to three significant digits, that's \( 5.62 \times 10^{-6} \)? Wait, no, wait, 5.25: let's do it directly. \( 10^{-5.25} \). Let's use a calculator: \( 10^{-5.25} = e^{-5.25 \times \ln 10} \approx e^{-5.25 \times 2.302585093} \approx e^{-12.0885715} \)? No, that's a mistake. Wait, no, \( \ln(10) \) is about 2.302585, so \( -5.25 \times 2.302585 \approx -12.088 \), but \( e^{-12.088} \) is a very small number, which is wrong. Wait, no! I messed up the exponent. Wait, \( 10^{-5.25} \) is not \( e^{-5.25 \ln 10} \), that's correct, but \( 10^{-5.25} \) is equal to \( 10^{-5} \times 10^{-0.25} \), which is \( 0.00001 \times 10^{-0.25} \). Wait, \( 10^{-5} \) is 0.00001, and \( 10^{-0.25} \) is approximately 0.5623, so 0.00001 * 0.5623 = 0.000005623, which is \( 5.623 \times 10^{-6} \). Yes, that's correct. So with three significant digits, it's \( 5.62 \times 10^{-6} \)? Wait, no, 5.623 rounded to three significant digits is 5.62? Wait, 5.623: the first three significant digits are 5, 6, 2, and the next digit is 3, which is less than 5, so we keep it 5.62. Wait, but let's check with a calculator. Let's compute \( 10^{-5.25} \). Using a calculator, \( 10^{-5.25} = 5.623413251903491 \times 10^{-6} \). So rounding to three significant digits, that's \( 5.62 \times 10^{-6} \)? Wait, no, 5.62341325... the first three significant digits are 5, 6, 2, and the fourth is 3, so we round down, so 5.62 × 10⁻⁶? Wait, no, 5.62341325 × 10⁻⁶, three significant digits: 5.62 (since the fourth digit is 3, which is less than 5, so we don't round up the third digit). Wait, but 5.62341325 is approximately 5.62 × 10⁻⁶ when rounded to three significant digits.

Wait, but let's do it again. The formula is \( [\text{H}^+] = 10^{-\text{pH}} \). So pH = 5.25, so \( [\text{H}^+] = 10^{-5.25} \). Let's calculate \( 10^{-5.25} \). Let's write -5.25 as -5 - 0.25, so \( 10^{-5.25} = 10^{-5} \times 10^{-0.25} \). \( 10^{-0.25} = 1 / 10^{0.25} \). \( 10^{0.25} = \sqrt[4]{10} \approx 1.77827941 \), so \( 10^{-0.25} \approx 0.562341325 \). Then, \( 10^{-5} \times 0.562341325 = 0.562341325 \times 10^{-5} = 5.62341325 \times 10^{-6} \). So with three significant digits, that's \( 5.62 \times 10^{-6} \)? Wait, no, 5.62341325, the first three significant digits are 5, 6, 2, and the next digit is 3, so we round down, so 5.62 × 10⁻⁶. Wait, but let's check with a calculator. If we use a calculator to compute \( 10^{-5.25} \), we get approximately 5.62341325 × 10⁻⁶, which is 5.62 × 10⁻⁶ when rounded to three significant digits.

Answer:

\( 5.62 \times 10^{-6} \) (entered as 5.62*10^-6)