What is the ph of 1.00 l of the buffer solution that contains 0.100 m hf and 0.120 m naf after 0.020 mol of hno₃ is added?
the of hf is 3.5×10⁻⁴, and so p = 3.46.
express your answer using two decimal places.

pH = 3,38

Explanation:

The reaction of HNO₃ with NaF is:

HNO₃ + NaF → HF + NaNO₃

If you have 1L of 0,120M NaF, moles of NaF are 0,120 moles. These moles reacts with 0,020 mol of HNO₃ to produce 0,020 moles of HF. That means the final concentrations of HF and NaF are:

HF: 0,100mol + 0,020 mol = 0,120 mol

NaF: 0,120mol - 0,020 mol = 0,100 mol

Now, to obtain the pH of this buffer (Weak acid + conjugate base) you can use  Henderson-Hasselblach formula:

pH = pka + log₁₀ [NaF] / [HF]

Replacing:

pH = 3.46 + log₁₀ [0,100] / [0,120]

pH = 3.38

I hope it helps!

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

because u need two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei to form isotopes.

so since they all have the same shape? i mean maybe because m& m and reeses are both chocolates