0.70 gg of hydrogen chloride (HClHCl) is dissolved in water to make 4.0 LL of solution. What is the pHpH of the resulting hydrochloric acid solution? Express the pH numerically to two decimal places.

The pH of the hydrochloric solution is 2.32

Explanation:

Step 1: Data given

Mass of HCl = 0.70 grams

Volume = 4.0 L

Molar mass of HCl = 36.46 g/mol

Step 2: Calculate moles HCl

Moles HCl = mass HCl / molar mass HCl

Moles HCl = 0.70 grams / 36.46g/mol

Moles HCl = 0.0192 moles

Step 3: Calculate molarity

Molarity HCl = moles HCl / volume

Molarity HCl = 0.0192 moles / 4.0 L

Molarity HCl = 0.0048 M

Step 4: Calculate molarity H+

For 1 mol HCl we have 1 mol H+

Molarity HCl = molarity H+ = 0.0048 M

Step 5: Calculate pH

pH = -log[H+]

pH = -log (0.0048)

pH = 2.32

The pH of the hydrochloric solution is 2.32

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol

can you attach the