Determine the pH of each of the following solutions., 3.6×10−2 M HI,9.23×10−2 M HClO4, a solution that is 4.0×10−2 M in HClO4 and 4.8×10−2 M in HCl, a solution that is 1.01% HCl by mass (Assume a density of 1.01 g/mL for the solution.)

please show work

## Answer

**Answer –** We are given the acids and need to
calculate the pH of each

**1)[**HI] = 3.6*10^{-2} M

We know the HI is the strong acid, so

[HI] = [H_{3}O^{+}] = 3.6*10^{-2} M

We know,

pH = -log [H_{3}O^{+}]

= -log 3.6*10^{-2} M

**= 1.44**

**2)[**HClO_{4}] = 9.23*10^{-2}
M

We know the HClO_{4} is the strong acid, so

[HClO_{4}] = [H_{3}O^{+}] =
9.23*10^{-2} M

We know,

**pH** = -log [H_{3}O^{+}]

= -log 9.23*10^{-2} M

= **1.03**

**3)** Solution [HClO_{4}] =
4.0*10^{-2} M and [HCl] = 9.23*10^{-2} M

We know the HClO_{4} and HCl both are the strong acid,
so

[HClO_{4}] = [H_{3}O^{+}] =
4.0*10^{-2} M

[HClO_{4}] = [H_{3}O^{+}] =
4.8*10^{-2} M

Total, [H_{3}O^{+}] = 4.0*10^{-2} M +
4.8*10^{-2} M

= 8.8*10^{-2} M

We know,

**pH =** -log [H_{3}O^{+}]

= -log 8.8*10^{-2} M

** = 1.05**

**4)** solution that is 1.01% HCl by mass, density
of solution = 1.01 g/mL

We assume 100 g of solution

So, mass of HCl = 1.01 g

volume of solution = 100 g / 1.01g.mL^{-1}

= 99.0 mL

= 0.099 L

Moles of HCl = 1.01 g / 36.456 g.mol^{-1}

= 0.0277 moles

[HCl] = 0.0277 moles / 0.099 L

= 0.280 M

We know,

HCl is the strong acid, so

[HClO_{4}] = [H_{3}O^{+}] = 0.280M

We know,

**pH** = -log [H_{3}O^{+}]

= -log 0.280 M

= **0.553**

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