Question

the approximate molar concentration of several chemicals are given. find the ph of each. use the calculator and round to the nearest tenth, if necessary. oven cleaner: h+=10^(-13) ph = water: h+=0.0000007 ph = blood: h+=0.00000004 ph = vinegar: h+=0.0063 ph =

Explanation:

Step1: Recall pH formula

The formula for pH is $pH = -\log[H^{+}]$.

Step2: Calculate pH of oven - cleaner

Given $[H^{+}]=10^{-13}$, then $pH = -\log(10^{-13})$. Since $-\log(10^{-13})=-(- 13)\log(10)$ and $\log(10) = 1$, so $pH = 13$.

Step3: Calculate pH of water

Given $[H^{+}]=0.0000007 = 7\times10^{-7}$. Then $pH=-\log(7\times10^{-7})= - (\log7+\log(10^{-7}))$. Since $\log(10^{-7})=-7$ and $\log7\approx0.845$, $pH\approx-(0.845 - 7)=6.2$.

Step4: Calculate pH of blood

Given $[H^{+}]=0.00000004 = 4\times10^{-8}$. Then $pH = -\log(4\times10^{-8})=-(\log4+\log(10^{-8}))$. Since $\log(10^{-8})=-8$ and $\log4\approx0.602$, $pH\approx-(0.602 - 8)=7.4$.

Step5: Calculate pH of vinegar

Given $[H^{+}]=0.0063 = 6.3\times10^{-3}$. Then $pH=-\log(6.3\times10^{-3})=-(\log6.3+\log(10^{-3}))$. Since $\log(10^{-3})=-3$ and $\log6.3\approx0.80$, $pH\approx-(0.80 - 3)=2.2$.

Answer:

Oven cleaner: 13
Water: 6.2
Blood: 7.4
Vinegar: 2.2