It is known that a finite dimensional Hopf algebra $H$ over a field $k$ is a Frobenius algebra. Thus there is an isomorphism $H \cong H^\ast$ of left $H$-modules.
Question: Is it possible to write down such an isomorphism entirely in terms of the Hopf algebra, i.e. by product, coproduct, unit, counit and involution ?
This isn't an answer but a lengthy comment.
The proofs for the $H$-modul ismorphism $H \cong H^\ast$ I know of use (some variant of) the $H$-module isomorphism $I_L(H^\ast) \otimes_k H \cong H^\ast$ where $$I_L(H^\ast) = \lbrace g \in H^\ast \mid \forall f\in H^\ast: f\ast g = f(1) \cdot g \rbrace$$ is the space of left integrals of $H^\ast$ ($f \ast g$ is the convolution). A dimension argument shows that $I_L(H^\ast)$ is one-dimensional. If $\lambda$ is a non-zero left integral then $I_L(H^\ast) = k\cdot \lambda \cong k$ yields $H \cong H^\ast$ as left $H$-modules. This is just the isomorphism described by darij.
The problem is that the proof doesn't yield a formula for such a $\lambda$. Choose a $k$-basis $\lbrace e_1=1,e_2,...,e_n\rbrace$ of $H$ and suppose $\Delta(e_k) = \sum_{i,j}d_{ij}^{(k)}\cdot e_i \otimes e_j$. Set $D^{(k)} = (d_{ij}^{(k)}-\delta_{i,1}\cdot \delta_{kj})_{i,j=1,\dots,n}$. Expressing $\lambda$ as linear combination of the dual basis then leads to the system of linear equations $D^{(k)} x = 0$ $(k=1,...,n)$. As pointed out by Mariano it has a unique solution (up to scalars). Again, this is no closed formula, but it can be used in practise to compute integrals.
