Question

module 8: homework assignment - chapter 14 exercise 14.74 - enhanced - with feedback missed this? watch kcv: the ph scale, we: calculating ph from h₃o⁺; read section 14.9. you can click on the review link to access the section in your etext. calculate h₃o⁺ for each solution given its ph part a ph = 1.26 express your answer using two significant figures. h₃o⁺ = 7.5·10⁻² you have already submitted this answer. enter a new answer. no credit lost. try again. submit previous answers request answer

Explanation:

Step1: Recall the pH formula

The formula that relates pH and the concentration of hydronium ions ($\ce{[H3O+]}$) is $pH = -\log_{10}(\ce{[H3O+]})$. We can rearrange this formula to solve for $\ce{[H3O+]}$: $\ce{[H3O+]} = 10^{-pH}$.

Step2: Substitute the given pH value

We are given that $pH = 1.26$. Substitute this value into the formula: $\ce{[H3O+]} = 10^{-1.26}$.

Step3: Calculate the value

First, calculate $10^{-1.26}$. We know that $10^{-1.26}=10^{-1 - 0.26}=10^{-1}\times10^{-0.26}$. $10^{-1} = 0.1$, and $10^{-0.26}\approx0.5495$. Then, multiply these two values: $0.1\times0.5495 = 0.05495$. Rounding to two significant figures, we get $0.055$ or $5.5\times10^{-2}$. Wait, wait, no, let's do the calculation more accurately. Using a calculator, $10^{-1.26}\approx5.5\times10^{-2}$? Wait, no, $pH = 1.26$, so $-\log_{10}(\ce{[H3O+]})=1.26$, so $\ce{[H3O+]} = 10^{-1.26}$. Let's compute $10^{-1.26}$:

We can write $1.26 = 1 + 0.26$, so $10^{-1.26}=10^{-1}\times10^{-0.26}$. $10^{-0.26}$: take the natural logarithm: $\ln(10^{-0.26})=-0.26\ln(10)\approx -0.26\times2.3026\approx -0.5987$, then $e^{-0.5987}\approx0.549$. Then $10^{-1}\times0.549 = 0.0549\approx5.5\times10^{-2}$ (two significant figures). Wait, but let's check with a calculator directly: $10^{-1.26}$. Let's use a calculator: $1.26$ negative, then $10^x$. So $10^{-1.26}\approx5.5\times10^{-2}$ M. Wait, but let's do it properly. Let's calculate $10^{-1.26}$:

$10^{-1.26} = 10^{-(1 + 0.26)} = 10^{-1} \times 10^{-0.26}$. $10^{-0.26}$: using logarithm properties, $\log_{10}(x)= -0.26\implies x = 10^{-0.26}$. Let's compute $10^{-0.26}$:

We know that $\log_{10}(0.05) = \log_{10}(5\times10^{-2}) = \log_{10}(5) - 2\approx0.6990 - 2 = -1.3010$, which is more negative than -0.26. Wait, no, I messed up. Wait, $pH = 1.26$ is an acidic solution, so $\ce{[H3O+]}$ should be between $10^{-1}$ (pH=1) and $10^{-2}$ (pH=2). Since pH=1.26 is between 1 and 2, $\ce{[H3O+]}$ is between $10^{-1}$ (0.1 M) and $10^{-2}$ (0.01 M). Let's calculate $10^{-1.26}$:

Using a calculator, $10^{-1.26} \approx 5.5\times10^{-2}$ M (5.5 times 10 to the power of -2). Wait, let's check with a calculator: 1.26, press +/- to get -1.26, then press 10^x. Let's do that: on a calculator, $10^{-1.26}\approx0.05495\approx5.5\times10^{-2}$ (two significant figures).

Wait, but the initial wrong answer in the image was $7.5\times10^{-2}$, which is incorrect. So the correct calculation is:

$\ce{[H3O+]} = 10^{-pH} = 10^{-1.26}\approx5.5\times10^{-2}$ M (two significant figures).

Wait, maybe I made a mistake earlier. Let's recalculate $10^{-1.26}$:

Using a calculator (let's use a scientific calculator):

$10^{-1.26} = e^{-1.26\ln(10)} \approx e^{-1.26\times2.302585093} \approx e^{-2.899257217} \approx 0.05495 \approx 5.5\times10^{-2}$ (two significant figures). Yes, that's correct. So the correct $\ce{[H3O+]}$ is approximately $5.5\times10^{-2}$ M.

Answer:

$\boxed{5.5\times10^{-2}}$ (or $0.055$)