The ph of a solution prepared by mixing 45.0 ml of 0.183 m koh and 35.0 ml of 0.145 m hcl is

12.6.

Explanation:

no. of millimoles of KOH = (MV)KOH = (0.183 M)(45.0 mL) = 8.235 mmol.

no. of millimoles of HCl = (MV)HCl = (0.145 M)(35.0 mL) = 5.075 mmol.

So,

[OH⁻] = [(no. of millimoles of KOH) - (no. of millimoles of HCl)] / (V total) = (8.235 mmol - 5.075 mmol) / (80.0 mL) = 0.395 M.

∵ pOH = -log[OH⁻]

∴ pOH = -log(0.395 M) = 1.4.

∵ pH + pOH = 14.

∴ pH = 14 - pOH = 14 - 1.4 = 12.6.

explanation:

venn diagram is pictorial representation with of circles showing the relation between two finite sets or things. the intersection part of two circles conveys the relation present in between those sets or things.

here in the venn diagram of sand and salt two circles represent sand and salt. but the area falling under the intersection of these circle represents the mixture of sand and salt.

so, in a venn diagram the mixture will be placed in a middle that is in between the intersection of two circles.

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